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Trigonometric Equations



General Solutions vs Restricted Solutions
Solving Basic Sine Equations
Solving Basic Cosine Equations
Solving Basic Tangent Equations
Equations Requiring Algebraic Manipulation
Equations Requiring Identity Substitution
Quadratic-Type Trigonometric Equations
Equations with Multiple Angles
Extraneous Solutions



Solving for Unknown Angles

A trigonometric equation is an equation in which the unknown appears inside a trigonometric function: sin⁡(x)=12\sin(x) = \frac{1}{2}, 2cos⁡2(x)−1=02\cos^2(x) - 1 = 0, tan⁡(3x)=−1\tan(3x) = -1. 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=π6x = \frac{\pi}{6} satisfies sin⁡(x)=12\sin(x) = \frac{1}{2}, but so does x=5π6x = \frac{5\pi}{6}, and so does x=π6+2πx = \frac{\pi}{6} + 2\pi, and x=5π6+2πx = \frac{5\pi}{6} + 2\pi, and every other angle obtained by adding integer multiples of 2π2\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∈Zn \in \mathbb{Z}, and the restricted solution, which lists only those solutions within a specified interval — typically [0,2π)[0, 2\pi) or [0°,360°)[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.



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 nn.

For example, sin⁡(x)=12\sin(x) = \frac{1}{2} has two solutions per period of 2π2\pi: x=π6x = \frac{\pi}{6} and x=5π6x = \frac{5\pi}{6}. The general solution is:

x=π6+2nπorx=5π6+2nπ,n∈Zx = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}


For tangent, which has period π\pi and only one solution per period, the general solution is more compact. If tan⁡(x)=1\tan(x) = 1, the principal solution is x=π4x = \frac{\pi}{4}, and the general solution is:

x=π4+nπ,n∈Zx = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}


A restricted solution lists only the solutions within a specified interval — most commonly [0,2π)[0, 2\pi) or [0°,360°)[0°, 360°). For sin⁡(x)=12\sin(x) = \frac{1}{2} on [0,2π)[0, 2\pi), the answer is simply x=π6x = \frac{\pi}{6} and x=5π6x = \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|a| < 1 have two solutions per period of 2π2\pi. When ∣a∣=1|a| = 1, there is one (at the extreme of the function). Tangent and cotangent equations have exactly one solution per period of π\pi. These counts multiply when the argument is a multiple angle — sin⁡(2x)=a\sin(2x) = a has up to four solutions per 2π2\pi interval, sin⁡(3x)=a\sin(3x) = a has up to six, and so on.

Solving Basic Sine Equations

    An equation of the form sin⁡(x)=a\sin(x) = a is solvable if and only if −1≤a≤1-1 \leq a \leq 1 — a consequence of the bounded range of sine. If ∣a∣>1|a| > 1, there is no solution.

    Step 1: Find the reference angle. Compute α=arcsin⁡(∣a∣)\alpha = \arcsin(|a|) — the acute angle whose sine equals ∣a∣|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.

  • a>0a > 0: solutions are in Quadrants I and II.
  • a<0a < 0: solutions are in Quadrants III and IV.
  • a=0a = 0: solutions are on the xx-axis (0,π,2π,…0, \pi, 2\pi, \ldots).

    • Step 3: Write the solutions within one period $[0, 2\pi)$.

  • x=αx = \alpha
  • x=π−αx = \pi - \alpha
  • x=π+αx = \pi + \alpha
  • x=2π−αx = 2\pi - \alpha

    • Select the two quadrants from Step 2.

    Step 4: Write the general solution. Add 2nπ2n\pi to each solution from Step 3.

    Example: Solve sin⁡(x)=−32\sin(x) = -\frac{\sqrt{3}}{2}.

    Reference angle: α=arcsin⁡(32)=π3\alpha = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}. Sine is negative → Quadrants III and IV:

    x=π+π3=4π3orx=2π−π3=5π3x = \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=4π3+2nπx = \frac{4\pi}{3} + 2n\pi or x=5π3+2nπx = \frac{5\pi}{3} + 2n\pi.

Solving Basic Cosine Equations

    The equation cos⁡(x)=a\cos(x) = a is solvable if and only if −1≤a≤1-1 \leq a \leq 1. The process parallels sine equations but with different quadrant assignments.

    Step 1: Reference angle. α=arccos⁡(∣a∣)\alpha = \arccos(|a|) — the angle in [0,π2][0, \frac{\pi}{2}] whose cosine equals ∣a∣|a|.

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

    Step 3: Solutions within $[0, 2\pi)$.

  • x=αx = \alpha
  • x=π−αx = \pi - \alpha
  • x=π+αx = \pi + \alpha
  • x=2π−αx = 2\pi - \alpha

    • Step 4: General solution. Add 2nπ2n\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=αx = \alpha and x=2π−α=−α+2πx = 2\pi - \alpha = -\alpha + 2\pi. In general solution form:

    x=±α+2nπ,n∈Zx = \pm\alpha + 2n\pi, \quad n \in \mathbb{Z}


    This compact form captures both solutions (the +α+\alpha branch and the −α-\alpha branch) in a single expression.

    Example: Solve cos⁡(x)=−12\cos(x) = -\frac{1}{2} on [0,2π)[0, 2\pi).

    Reference angle: α=arccos⁡(12)=π3\alpha = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}. Cosine is negative → Quadrants II and III:

    x=π−π3=2π3orx=π+π3=4π3x = \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\tan(x) = a has solutions for every real value of aa — 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. α=arctan⁡(a)\alpha = \arctan(a) — the unique angle in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) whose tangent equals aa. This is always well-defined, since arctangent accepts any real input.

Step 2: General solution. Because tangent has period π\pi and is one-to-one on each period interval, there is exactly one solution per period:

x=α+nπ,n∈Zx = \alpha + n\pi, \quad n \in \mathbb{Z}


No quadrant analysis is needed — the single principal value α\alpha, repeated at intervals of π\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)=3\tan(x) = \sqrt{3}.

Principal value: α=arctan⁡(3)=π3\alpha = \arctan(\sqrt{3}) = \frac{\pi}{3}.

General solution: x=π3+nπx = \frac{\pi}{3} + n\pi.

On [0,2π)[0, 2\pi): x=π3x = \frac{\pi}{3} (from n=0n = 0) and x=π3+π=4π3x = \frac{\pi}{3} + \pi = \frac{4\pi}{3} (from n=1n = 1).

Example: Solve tan⁡(x)=−1\tan(x) = -1.

Principal value: α=arctan⁡(−1)=−π4\alpha = \arctan(-1) = -\frac{\pi}{4}.

General solution: x=−π4+nπx = -\frac{\pi}{4} + n\pi.

On [0,2π)[0, 2\pi): x=−π4+π=3π4x = -\frac{\pi}{4} + \pi = \frac{3\pi}{4} and x=−π4+2π=7π4x = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4}.

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\sin(x) = a, cos⁡(x)=a\cos(x) = a, or tan⁡(x)=a\tan(x) = a.

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

sin⁡(x)cos⁡(x)=0⇒sin⁡(x)=0orcos⁡(x)=0\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 [0,2π)[0, 2\pi): sin⁡(x)=0\sin(x) = 0 gives x=0,πx = 0, \pi; cos⁡(x)=0\cos(x) = 0 gives x=π2,3π2x = \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⇒sin⁡(x)[cos⁡(x)+1]=0\sin(x)\cos(x) + \sin(x) = 0 \quad \Rightarrow \quad \sin(x)[\cos(x) + 1] = 0

This gives sin⁡(x)=0\sin(x) = 0 or cos⁡(x)=−1\cos(x) = -1. On [0,2π)[0, 2\pi): x=0,πx = 0, \pi from the first, and x=πx = \pi from the second. The combined solution set is {0,π}\{0, \pi\}.

A critical warning about division. It is tempting to "simplify" sin⁡(x)tan⁡(x)=sin⁡(x)\sin(x)\tan(x) = \sin(x) by dividing both sides by sin⁡(x)\sin(x). This discards the solutions where sin⁡(x)=0\sin(x) = 0. The correct approach is to rearrange to sin⁡(x)tan⁡(x)−sin⁡(x)=0\sin(x)\tan(x) - \sin(x) = 0, factor as sin⁡(x)[tan⁡(x)−1]=0\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 xx 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\sin^2(x) + \cos(x) = 1


Replace sin⁡2(x)\sin^2(x) with 1−cos⁡2(x)1 - \cos^2(x):

1−cos⁡2(x)+cos⁡(x)=1⇒−cos⁡2(x)+cos⁡(x)=01 - \cos^2(x) + \cos(x) = 1 \quad \Rightarrow \quad -\cos^2(x) + \cos(x) = 0


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


This gives cos⁡(x)=0\cos(x) = 0 or cos⁡(x)=1\cos(x) = 1. On [0,2π)[0, 2\pi): x=π2,3π2x = \frac{\pi}{2}, \frac{3\pi}{2} from the first, and x=0x = 0 from the second.

Another common scenario involves the identity sec⁡2(x)=1+tan⁡2(x)\sec^2(x) = 1 + \tan^2(x):

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


Replace sec⁡2(x)\sec^2(x):

1+tan⁡2(x)−tan⁡(x)=3⇒tan⁡2(x)−tan⁡(x)−2=01 + \tan^2(x) - \tan(x) = 3 \quad \Rightarrow \quad \tan^2(x) - \tan(x) - 2 = 0


This is a quadratic in tan⁡(x)\tan(x), solvable by factoring or the quadratic formula.

The double angle formulas are another frequent source of substitution. An equation containing cos⁡(2x)\cos(2x) alongside sin⁡(x)\sin(x) can be converted using cos⁡(2x)=1−2sin⁡2(x)\cos(2x) = 1 - 2\sin^2(x), producing a quadratic in sin⁡(x)\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⁡(2θ)\cos(2\theta) — cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta, 2cos⁡2θ−12\cos^2\theta - 1, and 1−2sin⁡2θ1 - 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, 2sin⁡2(x)−sin⁡(x)−1=02\sin^2(x) - \sin(x) - 1 = 0 — is solved by treating the trigonometric expression as a temporary variable.

    Let u=sin⁡(x)u = \sin(x). The equation becomes:

    2u2−u−1=02u^2 - u - 1 = 0


    Factor: (2u+1)(u−1)=0(2u + 1)(u - 1) = 0, giving u=−12u = -\frac{1}{2} or u=1u = 1.

    Now solve each:

  • sin⁡(x)=−12\sin(x) = -\frac{1}{2}: on [0,2π)[0, 2\pi), x=7π6x = \frac{7\pi}{6} and x=11π6x = \frac{11\pi}{6}.
  • sin⁡(x)=1\sin(x) = 1: on [0,2π)[0, 2\pi), x=π2x = \frac{\pi}{2}.

    • The complete solution set on [0,2π)[0, 2\pi) is {π2,7π6,11π6}\left\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\right\}.

    When factoring is not obvious, the quadratic formula applies:

    u=−b±b2−4ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}


    After finding the values of uu, check whether each falls within the range of the relevant function. For sine and cosine, reject any ∣u∣>1|u| > 1 — these correspond to no solution. For tangent, any real value of uu 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)\sin(x) and cos⁡(x)\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\sin(2x) = a, cos⁡(3x)=a\cos(3x) = a, tan⁡(x2)=a\tan\left(\frac{x}{2}\right) = a — the solving process adds an extra step: solve for the inner argument first, then recover xx.

    Example: Solve sin⁡(2x)=32\sin(2x) = \frac{\sqrt{3}}{2} on [0,2π)[0, 2\pi).

    Let u=2xu = 2x. The equation sin⁡(u)=32\sin(u) = \frac{\sqrt{3}}{2} has solutions:

    u=π3+2nπoru=2π3+2nπu = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad u = \frac{2\pi}{3} + 2n\pi


    Since u=2xu = 2x, divide by 2:

    x=π6+nπorx=π3+nπx = \frac{\pi}{6} + n\pi \quad \text{or} \quad x = \frac{\pi}{3} + n\pi


    Now find which values fall in [0,2π)[0, 2\pi):

  • x=π6x = \frac{\pi}{6} (n=0n=0), x=π6+π=7π6x = \frac{\pi}{6} + \pi = \frac{7\pi}{6} (n=1n=1)
  • x=π3x = \frac{\pi}{3} (n=0n=0), x=π3+π=4π3x = \frac{\pi}{3} + \pi = \frac{4\pi}{3} (n=1n=1)

    • Four solutions: {π6,π3,7π6,4π3}\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 2π2\pi interval. A factor of 3 would triple it. In general, sin⁡(nx)=a\sin(nx) = a on [0,2π)[0, 2\pi) can produce up to 2n2n solutions (for ∣a∣<1|a| < 1). This multiplication is the most common source of missed solutions — students who solve for uu but forget to generate all valid xx values within the target interval.

    When the interval for xx is [0,2π)[0, 2\pi) and the argument is 2x2x, the effective interval for uu is [0,4π)[0, 4\pi). For 3x3x, it becomes [0,6π)[0, 6\pi). Always determine the uu-interval before listing solutions — otherwise, solutions from later periods will be overlooked.

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

    Square both sides:

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


    Using sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1:

    1+2sin⁡(x)cos⁡(x)=1⇒2sin⁡(x)cos⁡(x)=0⇒sin⁡(2x)=01 + 2\sin(x)\cos(x) = 1 \quad \Rightarrow \quad 2\sin(x)\cos(x) = 0 \quad \Rightarrow \quad \sin(2x) = 0


    So 2x=nπ2x = n\pi, giving x=nπ2x = \frac{n\pi}{2}. On [0,2π)[0, 2\pi): x=0,π2,π,3π2x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}.

    Now check each in the original equation sin⁡(x)+cos⁡(x)=1\sin(x) + \cos(x) = 1:

  • x=0x = 0: 0+1=10 + 1 = 1 ✓
  • x=π2x = \frac{\pi}{2}: 1+0=11 + 0 = 1 ✓
  • x=πx = \pi: 0+(−1)=−1≠10 + (-1) = -1 \neq 1 ✗
  • x=3π2x = \frac{3\pi}{2}: −1+0=−1≠1-1 + 0 = -1 \neq 1 ✗

    • Only x=0x = 0 and x=π2x = \frac{\pi}{2} are valid. The squaring step introduced two extraneous solutions where sin⁡(x)+cos⁡(x)=−1\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 aa from −a-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.