. Unlike identities, which hold for every angle, an equation is satisfied only by specific values — and the challenge is to find them all. The word "all" is critical, because periodicity guarantees that if one solution exists, infinitely many do. The angle

x = \frac{\pi}{6}

satisfies

\sin(x) = \frac{1}{2}

, but so does

x = \frac{5\pi}{6}

, and so does

x = \frac{\pi}{6} + 2\pi

, and

x = \frac{5\pi}{6} + 2\pi

, and every other angle obtained by adding integer multiples of

2\pi

to either of these.

Managing this infinity is what distinguishes trigonometric equation solving from solving polynomial or rational equations. Two formats are standard: the general solution, which captures every solution using a parameter

n \in \mathbb{Z}

, and the restricted solution, which lists only those solutions within a specified interval — typically

[0, 2\pi)

[0°, 360°)

. The tools required include the unit circle (for reading solutions geometrically), inverse trigonometric functions (for computing principal values), identities (for rewriting multi-function equations), and formulas (for handling double angles, half angles, and other compound arguments). The properties of the trigonometric functions — particularly periodicity and boundedness — govern when solutions exist and how many there are.

Key Terms

Periodic Function— a function that repeats at regular intervals, producing infinitely many solutions

Reference Angle— the acute angle to the

x

-axis, used to identify solutions by quadrant

Inverse Trigonometric Function— returns the angle for a given function value

Supplementary Angles— angles summing to

180°

, linking sine solutions across quadrants

Quadrantal Angles— boundary angles where some functions are undefined or extreme

Formulas Used

General Solution - Sine Equation—

\sin\theta = k \Rightarrow \theta = (-1)^n \arcsin k + n\pi

General Solution - Cosine Equation—

\cos\theta = k \Rightarrow \theta = \pm\arccos k + 2n\pi

General Solution - Tangent Equation—

\tan\theta = k \Rightarrow \theta = \arctan k + n\pi

See All Trigonometry Definitions →

General Solutions vs Restricted Solutions

Because trigonometric functions are periodic, any equation that has at least one solution has infinitely many. The general solution captures all of them in a compact expression parameterized by an integer

n

.

For example,

\sin(x) = \frac{1}{2}

has two solutions per period of

2\pi

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

. The general solution is:

x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}

The condition

n \in \mathbb{Z}

reads '

n

ranges over the integers' —

\mathbb{Z}

is one of the standard set symbols that appears in every general solution.

For tangent, which has period

\pi

and only one solution per period, the general solution is more compact. If

\tan(x) = 1

, the principal solution is

x = \frac{\pi}{4}

, and the general solution is:

x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}

A restricted solution lists only the solutions within a specified interval — most commonly

[0, 2\pi)

[0°, 360°)

. For

\sin(x) = \frac{1}{2}

[0, 2\pi)

, the answer is simply

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

.

The problem statement determines which format is required. When no interval is specified and the problem says "find all solutions" or "solve," a general solution is expected. When an interval is given, only those solutions falling within it should be listed.

The number of solutions per period depends on the function and the value. Sine and cosine equations with

|a| < 1

have two solutions per period of

2\pi

. When

|a| = 1

, there is one (at the extreme of the function). Tangent and cotangent equations have exactly one solution per period of

\pi

. These counts multiply when the argument is a multiple angle —

\sin(2x) = a

has up to four solutions per

2\pi

interval,

\sin(3x) = a

has up to six, and so on.

y = sin x repeats every 2π

Any horizontal line between −1 and 1 cuts this wave twice in every period, and the cuts recur every 2π in both directions: that is why one solution always brings infinitely many with it, and why the general solution carries an integer n. The marker at 60° is one of the crossings for the line y = √3/2. Follow the wave through several periods on the trigonometric functions graphs.

Restricting to an interval simply selects which of these infinitely many crossings to report.

Solving Basic Sine Equations

An equation of the form

\sin(x) = a

is solvable if and only if

-1 \leq a \leq 1

— a consequence of the bounded range of sine. If

|a| > 1

, there is no solution.

Step 1: Find the reference angle. Compute

\alpha = \arcsin(|a|)

— the acute angle whose sine equals

|a|

. For standard values, this comes from the unit circle or special triangles. For non-standard values, use a calculator with the inverse sine function.

Step 2: Determine the quadrants. Sine is positive in Quadrants I and II, negative in Quadrants III and IV.

If $a > 0$ : solutions are in Quadrants I and II.
If $a < 0$ : solutions are in Quadrants III and IV.
If $a = 0$ : solutions are on the $x$ -axis ( $0, \pi, 2\pi, \ldots$ ).

Step 3: Write the solutions within one period.

Quadrant I: $x = \alpha$
Quadrant II: $x = \pi - \alpha$
Quadrant III: $x = \pi + \alpha$
Quadrant IV: $x = 2\pi - \alpha$

Select the two quadrants from Step 2.

Step 4: Write the general solution. Add

2n\pi

to each solution from Step 3. Equivalently, both branches can be collapsed into a single expression using an alternating sign:

General Solution - Sine Equation

\sin\theta = k \implies \theta = (-1)^n \arcsin k + n\pi

Learn more about this formula: General Solution - Sine Equation →

Browse all trigonometry formulas →

Example: Solve

\sin(x) = -\frac{\sqrt{3}}{2}

Reference angle:

\alpha = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}

. Sine is negative → Quadrants III and IV:

x = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \quad \text{or} \quad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}

General solution:

x = \frac{4\pi}{3} + 2n\pi

x = \frac{5\pi}{3} + 2n\pi

sin 30° = sin 150° = 1/2

Both hover boxes show the same y-coordinate, 1/2: the two points sit at the same height on opposite sides of the y-axis, one at the reference angle 30° in Quadrant I and one at 180° − 30° in Quadrant II. Those are the two solutions of sin x = 1/2 in one turn of the circle; every other solution is one of these plus full turns. Hover the mirrored pair for any height on the unit circle visualizer.

The sign of a only decides which pair of quadrants hosts the two points; the mirror relationship is the same.

Solving Basic Cosine Equations

The equation

\cos(x) = a

is solvable if and only if

-1 \leq a \leq 1

. The process parallels sine equations but with different quadrant assignments.

Step 1: Reference angle.

\alpha = \arccos(|a|)

— the angle in

[0, \frac{\pi}{2}]

whose cosine equals

|a|

Step 2: Quadrants. Cosine is positive in Quadrants I and IV, negative in Quadrants II and III.

Step 3: Solutions within one period.

Quadrant I: $x = \alpha$
Quadrant II: $x = \pi - \alpha$
Quadrant III: $x = \pi + \alpha$
Quadrant IV: $x = 2\pi - \alpha$

Step 4: General solution. Add

2n\pi

to each.

A cleaner notation is available for cosine because of its even symmetry. The two solutions within one period can be written as

x = \alpha

and

x = 2\pi - \alpha = -\alpha + 2\pi

. In general solution form:

General Solution - Cosine Equation

\cos\theta = k \implies \theta = \pm\arccos k + 2n\pi

Learn more about this formula: General Solution - Cosine Equation →

Browse all trigonometry formulas →

This compact form captures both solutions (the

+\alpha

branch and the

-\alpha

branch) in a single expression.

Example: Solve

\cos(x) = -\frac{1}{2}

[0, 2\pi)

Reference angle:

\alpha = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}

. Cosine is negative → Quadrants II and III:

x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \quad \text{or} \quad x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}

Solving Basic Tangent Equations

The equation

\tan(x) = a

has solutions for every real value of

a

— tangent's range is

(-\infty, \infty)

, so boundedness is never a constraint. This makes tangent equations structurally simpler than sine or cosine equations.

Step 1: Principal value.

\alpha = \arctan(a)

— the unique angle in

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

whose tangent equals

a

. This is always well-defined, since arctangent accepts any real input.

Step 2: General solution. Because tangent has period

\pi

and is one-to-one on each period interval, there is exactly one solution per period:

General Solution - Tangent Equation

\tan\theta = k \implies \theta = \arctan k + n\pi

Learn more about this formula: General Solution - Tangent Equation →

Browse all trigonometry formulas →

No quadrant analysis is needed — the single principal value

\alpha

, repeated at intervals of

\pi

, captures every solution. This is more compact than the sine and cosine cases, where two solutions per period require two separate expressions.

Example: Solve

\tan(x) = \sqrt{3}

.

Principal value:

\alpha = \arctan(\sqrt{3}) = \frac{\pi}{3}

.

General solution:

x = \frac{\pi}{3} + n\pi

.

On

[0, 2\pi)

x = \frac{\pi}{3}

(from

n = 0

) and

x = \frac{\pi}{3} + \pi = \frac{4\pi}{3}

(from

n = 1

).

Example: Solve

\tan(x) = -1

.

Principal value:

\alpha = \arctan(-1) = -\frac{\pi}{4}

.

General solution:

x = -\frac{\pi}{4} + n\pi

.

On

[0, 2\pi)

x = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}

and

x = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4}

Equation	Range constraint on a	Reference / principal angle	Solutions on [0, 2π)	General solution
sin x = a	\|a\| ≤ 1	α = arcsin\|a\|	a > 0: Q I, II → α, π − α a < 0: Q III, IV → π + α, 2π − α	x = α + 2nπ or x = π − α + 2nπ
cos x = a	\|a\| ≤ 1	α = arccos\|a\|	a > 0: Q I, IV → α, 2π − α a < 0: Q II, III → π − α, π + α	x = ±α + 2nπ
tan x = a	any real a	α = arctan a (∈ (−π ⁄ 2, π ⁄ 2))	α (mod π in [0, 2π)) and α + π — one per period	x = α + nπ

y = tan x: one crossing per branch

Every branch climbs through all real values exactly once, so a horizontal line at any height a meets the curve once per branch and the branches are π apart: tan x = a has exactly one solution per period of π, whatever a is. The marker at 60° is the crossing for a = √3. Pick any height and count the crossings on the trigonometric functions graphs.

The shorter period is why the general solution for tangent adds nπ rather than 2nπ.

Solution Set Notation

Notation

Solution Set Notation

Three devices carry every answer on this page: the integer-parameterized family, the sign that marches with the counter, and the plus-minus that cosine alone has earned.

\arcsin

\arccos

\arctan

and the

\sin^{-1}

collision belong to inverse functions; the qualifier

n \in \mathbb{Z}

and the step families come from function notation.

x = \frac{\pi}{6} + 2n\pi

x equals pi over six, plus two n pi

Solution-listing by parameter: one formula names infinitely many answers, one per integer

n

— the family notation pointed at solutions instead of exclusions. Two branches join with "or", as General Solutions vs Restricted Solutions above sets up.

CasesThe step matches the period —

+\,2n\pi

for sine and cosine,

+\,n\pi

for tangent's shorter cycle; a restricted answer trades the parameter for an interval, "on

[0, 2\pi)

", and lists members instead.

Do not confuseA single answer plus decoration.

n

runs over all of

\mathbb{Z}

, negatives included — expanding only

n = 0, 1, 2, \ldots

silently loses half the family.

(-1)^n \arcsin k + n\pi

minus one to the n, arcsine k, plus n pi

A sign that marches with the counter: even

n

keeps

+\alpha

, odd

n

flips to

-\alpha

— collapsing sine's two branches per period into one formula, the callout of Solving Basic Sine Equations above.

CasesThe alternating device is general-purpose in sequences and series; here it is welded to

n\pi

, so the sign flip and the half-turn step travel together — one counter drives both.

Do not confuse

-1^n

. Without parentheses the exponent grips only the

1

, and

-1^n = -1

for every

n

— the parentheses are the entire mechanism.

\pm\arccos k + 2n\pi

plus or minus arccos k, plus two n pi

Two branches in one glyph:

+\alpha

and

-\alpha

, legal for cosine because of its even symmetry — the cleaner form of Solving Basic Cosine Equations above. The mark itself belongs to roots and radicals; here it enumerates angle branches.

CasesExpanding comes before restricting: on

[0, 2\pi)

the

-\alpha

branch surfaces as

2\pi - \alpha

, its coterminal representative — the wrap idiom doing the bookkeeping.

Do not confuseA sine option.

\pm

does not transfer to sine — its branches sit at

\alpha

and

\pi - \alpha

, not

\pm\alpha

; that asymmetry is exactly why sine needs the

(-1)^n

device instead.

All trigonometry symbols →↑ Trigonometry

Equations Requiring Algebraic Manipulation

Many trigonometric equations cannot be solved by direct inversion — they require algebraic manipulation first. The goal is to reduce the equation to one or more basic equations of the form

\sin(x) = a

\cos(x) = a

, or

\tan(x) = a

.

Factoring. When a product equals zero, at least one factor must be zero.

\sin(x)\cos(x) = 0 \quad \Rightarrow \quad \sin(x) = 0 \quad \text{or} \quad \cos(x) = 0

These are two separate basic equations, each solved independently. On

[0, 2\pi)

\sin(x) = 0

gives

x = 0, \pi

;

\cos(x) = 0

gives

x = \frac{\pi}{2}, \frac{3\pi}{2}

. All four are solutions.

Grouping and factoring. When terms share a common factor:

\sin(x)\cos(x) + \sin(x) = 0 \quad \Rightarrow \quad \sin(x)[\cos(x) + 1] = 0

This gives

\sin(x) = 0

\cos(x) = -1

. On

[0, 2\pi)

x = 0, \pi

from the first, and

x = \pi

from the second. The combined solution set is

\{0, \pi\}

.

A critical warning about division. It is tempting to "simplify"

\sin(x)\tan(x) = \sin(x)

by dividing both sides by

\sin(x)

. This discards the solutions where

\sin(x) = 0

. The correct approach is to rearrange to

\sin(x)\tan(x) - \sin(x) = 0

, factor as

\sin(x)[\tan(x) - 1] = 0

, and solve each factor. Dividing by a trigonometric expression is valid only if that expression is provably nonzero — and in an equation where any value of

x

might be a solution, this condition is rarely guaranteed.

Equations Requiring Identity Substitution

When an equation involves more than one trigonometric function, a direct solution is usually impossible. The strategy is to use an identity to rewrite the equation in terms of a single function, then solve.

The Pythagorean identity is the most common tool. Consider:

\sin^2(x) + \cos(x) = 1

Replace

\sin^2(x)

with

1 - \cos^2(x)

1 - \cos^2(x) + \cos(x) = 1 \quad \Rightarrow \quad -\cos^2(x) + \cos(x) = 0

\cos(x)[1 - \cos(x)] = 0

This gives

\cos(x) = 0

\cos(x) = 1

. On

[0, 2\pi)

x = \frac{\pi}{2}, \frac{3\pi}{2}

from the first, and

x = 0

from the second.

Another common scenario involves the identity

\sec^2(x) = 1 + \tan^2(x)

\sec^2(x) - \tan(x) = 3

Replace

\sec^2(x)

1 + \tan^2(x) - \tan(x) = 3 \quad \Rightarrow \quad \tan^2(x) - \tan(x) - 2 = 0

This is a quadratic in

\tan(x)

, solvable by factoring or the quadratic formula.

The double angle formulas are another frequent source of substitution. An equation containing

\cos(2x)

alongside

\sin(x)

can be converted using

\cos(2x) = 1 - 2\sin^2(x)

, producing a quadratic in

\sin(x)

. The choice of which identity to apply depends on which functions are present and which substitution produces a single-variable expression.

Not every identity substitution leads to a cleaner equation. If one substitution creates a more complicated expression, try a different form. The three versions of

\cos(2\theta)

—

\cos^2\theta - \sin^2\theta

2\cos^2\theta - 1

, and

1 - 2\sin^2\theta

— each serve different purposes depending on whether the target variable is sine or cosine.

Quadratic-Type Trigonometric Equations

An equation that is quadratic in a trigonometric function — for example,

2\sin^2(x) - \sin(x) - 1 = 0

— is solved by treating the trigonometric expression as a temporary variable.

Let

u = \sin(x)

. The equation becomes:

2u^2 - u - 1 = 0

Factor:

(2u + 1)(u - 1) = 0

, giving

u = -\frac{1}{2}

u = 1

Now solve each:

$\sin(x) = -\frac{1}{2}$ : on $[0, 2\pi)$ , $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$ .
$\sin(x) = 1$ : on $[0, 2\pi)$ , $x = \frac{\pi}{2}$ .

The complete solution set on

[0, 2\pi)

\left\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\}

When factoring is not obvious, the quadratic formula applies:

u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

After finding the values of

u

, check whether each falls within the range of the relevant function. For sine and cosine, reject any

|u| > 1

— these correspond to no solution. For tangent, any real value of

u

is valid. This check is a consequence of boundedness and prevents reporting nonexistent solutions.

Quadratic-type equations often arise after identity substitution. An equation that initially involves both

\sin(x)

and

\cos(x)

may reduce to a quadratic in one function after a Pythagorean substitution. The two-step process — identity substitution followed by quadratic solving — is one of the most common patterns in trigonometric equation solving.

Equations with Multiple Angles

When the argument of the trigonometric function is a multiple of the variable —

\sin(2x) = a

\cos(3x) = a

\tan\left(\frac{x}{2}\right) = a

— the solving process adds an extra step: solve for the inner argument first, then recover

x

Example: Solve

\sin(2x) = \frac{\sqrt{3}}{2}

[0, 2\pi)

Let

u = 2x

. The equation

\sin(u) = \frac{\sqrt{3}}{2}

has solutions:

u = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad u = \frac{2\pi}{3} + 2n\pi

Since

u = 2x

, divide by 2:

x = \frac{\pi}{6} + n\pi \quad \text{or} \quad x = \frac{\pi}{3} + n\pi

Now find which values fall in

[0, 2\pi)

$x = \frac{\pi}{6}$ ( $n=0$ ), $x = \frac{\pi}{6} + \pi = \frac{7\pi}{6}$ ( $n=1$ )
$x = \frac{\pi}{3}$ ( $n=0$ ), $x = \frac{\pi}{3} + \pi = \frac{4\pi}{3}$ ( $n=1$ )

Four solutions:

\left\{\frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}\right\}

The critical point: the factor of 2 in the argument doubles the number of solutions within a

2\pi

interval. A factor of 3 would triple it. In general,

\sin(nx) = a

[0, 2\pi)

can produce up to

2n

solutions (for

|a| < 1

). This multiplication is the most common source of missed solutions — students who solve for

u

but forget to generate all valid

x

values within the target interval.

When the interval for

x

[0, 2\pi)

and the argument is

2x

, the effective interval for

u

[0, 4\pi)

. For

3x

, it becomes

[0, 6\pi)

. Always determine the

u

-interval before listing solutions — otherwise, solutions from later periods will be overlooked.

Same line, same interval, twice the crossings

Doubling the angle squeezes two full cycles into the interval that held one, so the same horizontal line is crossed twice as often. The four marked points are exactly the four solutions listed above — and the two a hurried solver finds are only the first half of them.

Widening the interval for

u

[0, 4\pi)

before solving is what guarantees none are missed.

Extraneous Solutions

Certain solving techniques — most notably squaring both sides — can produce values that satisfy the transformed equation but not the original. These are extraneous solutions, and they must be identified and discarded.

Example: Solve

\sin(x) + \cos(x) = 1

Square both sides:

\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = 1

Using

\sin^2(x) + \cos^2(x) = 1

1 + 2\sin(x)\cos(x) = 1 \quad \Rightarrow \quad 2\sin(x)\cos(x) = 0 \quad \Rightarrow \quad \sin(2x) = 0

2x = n\pi

, giving

x = \frac{n\pi}{2}

. On

[0, 2\pi)

x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}

Now check each in the original equation

\sin(x) + \cos(x) = 1

$x = 0$ : $0 + 1 = 1$ ✓
$x = \frac{\pi}{2}$ : $1 + 0 = 1$ ✓
$x = \pi$ : $0 + (-1) = -1 \neq 1$ ✗
$x = \frac{3\pi}{2}$ : $-1 + 0 = -1 \neq 1$ ✗

Only

x = 0

and

x = \frac{\pi}{2}

are valid. The squaring step introduced two extraneous solutions where

\sin(x) + \cos(x) = -1

— the negative square root that squaring fails to distinguish from the positive.

Extraneous solutions arise whenever a non-reversible operation is applied:

Squaring does not distinguish $a$ from $-a$ .
Multiplying by a variable expression may introduce solutions where that expression is zero.
Certain identity substitutions may change the effective domain.

The safeguard is straightforward: always substitute solutions back into the original equation. This verification step is not optional — it is an integral part of the solving process whenever squaring or other non-reversible steps have been used.

Operation	Why it can introduce extraneous solutions	How to handle
Squaring both sides	cannot distinguish a from −a (both square to a²)	substitute every candidate back into the original equation
Multiplying by a variable expression	adds solutions where that expression equals 0	discard any candidate that makes the multiplier 0
Certain identity substitutions	may broaden the domain or change the equation's effective form	verify candidates in the original equation, not the transformed one

Summary: Approach by Equation Form

Every trigonometric equation eventually reduces to a basic equation

\sin x = a

\cos x = a

, or

\tan x = a

— what differs is the path. The capstone table below collects the equation forms encountered above, the first move that gets each one onto a productive track, and what to do once the reduction is complete.

Equation form	First move	Result	Then solve as
Basic: sin/cos/tan x = a	check \|a\| against the function's range; find reference angle	principal angle α	apply the quadrant rule, then write the general solution
Product equals 0 (factored)	set each factor to 0 separately	several basic equations	solve each, then union the solution sets
Quadratic in a single function	substitute u = sin x (or cos x, tan x); factor or apply the quadratic formula	values of u	reject u values outside the function's range, then solve each basic equation
Two or more functions mixed	apply a Pythagorean, double-angle, or other identity	single-function equation	basic or quadratic in that function
Multiple angle: sin/cos/tan(nx) = a	let u = nx and solve sin/cos/tan u = a first	general solution in u	divide by n, generate every x in the target interval (up to 2n solutions on [0, 2π))
Equation requiring squaring	square (only when needed), reduce to a basic or quadratic form	candidate solutions	substitute each candidate into the original to discard extraneous values

Trigonometric Equations FAQ

Why does a trigonometric equation have infinitely many solutions?

Because the functions repeat. Once one angle satisfies the equation, adding any whole number of periods gives another angle that satisfies it too. The general solution captures the whole family in one expression using an integer parameter, while a restricted solution lists only those falling inside a stated interval.Read more →

What is the general solution of sin x = a?

It is x = (−1)ⁿα + nπ, where α is the principal angle and n runs over all integers. The alternating sign is needed because sine's two solutions per cycle sit at α and π − α, which are not symmetric about zero. Cosine, whose solutions are symmetric, uses a plus-or-minus instead.Read more →

How do you solve a quadratic trigonometric equation?

Substitute a single letter for the trigonometric expression, solve the resulting quadratic by factoring or formula, then substitute back and solve each simple equation separately. Discard any root outside the range of the function, since values above 1 or below −1 are impossible for sine and cosine.Read more →

Why does squaring both sides create extra solutions?

Squaring destroys sign information, so an equation that failed because two sides had opposite signs will pass once both are squared. The extra roots satisfy the squared version but not the original. Every solution must therefore be substituted back into the original equation and discarded if it does not check out.Read more →

When does a trigonometric equation have no solution?

When it demands a value the function cannot produce. Sine and cosine never leave the interval from −1 to 1, so an equation like cos x = 2 has no solution at all. Tangent has no such bound and can equal any real number, so equations in tangent alone always have solutions.Read more →