Finding Intervals Where Trigonometric Conditions Hold
A trigonometric equation asks: at which angles does a function hit a specific value? A trigonometric inequality asks a broader question: over which intervals does a function stay above, below, or between given bounds? The answer is not a set of isolated angles but a union of intervals — continuous stretches of the real line where the condition is satisfied. And because the trigonometric functions are periodic, these intervals repeat.
Two methods dominate. The graphical approach plots the function and a horizontal boundary line, then reads off the regions where the curve lies on the correct side. The unit circle approach identifies the boundary angles (the solutions of the corresponding equation) and determines which arc of the circle satisfies the inequality by examining the sign of the function along that arc. Both methods yield the same result; the choice depends on whether a visual or geometric perspective is more natural for the problem at hand. In either case, the underlying logic draws on the same properties — periodicity, boundedness, continuity — and the same evaluation skills used in equation solving.
Solutions as Intervals
The solution to a trigonometric inequality is fundamentally different from the solution to an equation. An equation like sin(x)=21 produces isolated points: x=6π,65π, plus their periodic repetitions. An inequality like sin(x)>21 produces intervals: the entire stretch of angles between 6π and 65π (exclusive), plus periodic repetitions.
This shift from points to intervals is a consequence of continuity. Since sine is continuous, if sin(a)>21 and sin(b)>21 and there is no point between a and b where sin(x)=21, then sin(x)>21 for every x between a and b. The boundary points — the solutions of the corresponding equation — are where the function crosses the threshold, and the intervals between them are where the inequality holds.
On a restricted interval like [0,2π), the solution is a finite union of intervals. In general solution form, the pattern repeats: each interval shifts by 2π (for sine, cosine, cosecant, secant) or π (for tangent, cotangent) to produce the complete solution set.
Strict inequalities (>, <) produce open intervals at the boundary points — the boundary is not included. Non-strict inequalities (≥, ≤) produce closed intervals at boundary points — the boundary is included (provided the function is defined there). At vertical asymptotes, the interval is always open, since the function is undefined.
The Graphical Method
The graphical method is the most intuitive approach to trigonometric inequalities. It reduces the problem to a visual reading.
Step 1: Graph the function. Sketch or visualize y=sin(x) (or whichever function appears in the inequality) over the relevant interval. The key features — zeros, extrema, asymptotes, period — should be familiar from the study of graphs.
Step 2: Draw the boundary line. Plot the horizontal line y=a, where a is the value on the right side of the inequality.
Step 3: Identify the intersection points. These are the solutions of the corresponding equation (sin(x)=a), which can be found using the techniques from equation solving.
Step 4: Read the solution intervals. For sin(x)>a: the intervals where the curve lies above the horizontal line. For sin(x)<a: the intervals where the curve lies below.
Example: Solve cos(x)>21 on [0,2π).
The equation cos(x)=21 has solutions x=3π and x=35π on this interval. The cosine graph starts at 1 (above 21), drops to 21 at 3π, continues below 21 through the middle of the interval, rises back to 21 at 35π, and returns to 1 at 2π. The curve is above the line y=21 on:
x∈[0,3π)∪(35π,2π)
The endpoints 3π and 35π are excluded because the inequality is strict (>, not ≥). The endpoint 0 is included because the interval [0,2π) starts there and cos(0)=1>21.
The graphical method is especially useful for transformed functions — 2sin(3x)−1>0 is harder to handle algebraically, but sketching the transformed graph and reading off the solution intervals is straightforward once the transformations are understood.
The Unit Circle Method
The unit circle provides a geometric alternative to the graphical method, particularly effective for basic inequalities involving sin(x) and cos(x).
For sine inequalities: since sinθ is the y-coordinate on the unit circle, the inequality sin(x)>a asks: for which angles is the y-coordinate above the line y=a?
Draw the horizontal line y=a across the unit circle. It intersects the circle at two points (assuming ∣a∣<1), dividing the circle into two arcs. The arc where points have y-coordinates above a is the solution. Reading off the angles at the intersection points gives the interval boundaries.
For cosine inequalities: since cosθ is the x-coordinate, the inequality cos(x)>a asks: for which angles is the x-coordinate to the right of the vertical line x=a?
Draw the vertical line x=a across the unit circle. The arc to the right of this line gives the solution.
Example: Solve sin(x)≥22 on [0,2π).
On the unit circle, the horizontal line y=22 intersects the circle at angles 4π and 43π. The arc from 4π to 43π (going counterclockwise through the top of the circle) lies above this line. Since the inequality includes equality (≥), the endpoints are included:
x∈[4π,43π]
For tangent inequalities: the unit circle method is less direct, since tangent does not correspond to a single coordinate. One approach is to interpret tanθ as the slope of the terminal side and identify which slopes satisfy the inequality. Alternatively, convert to cosθsinθ and reason about the signs and magnitudes separately. In practice, the graphical method is often simpler for tangent inequalities.
Solving Sine Inequalities
The full procedure for solving sin(x)>a (or ≥, <, ≤) combines the boundary equation with interval determination.
Case $|a| < 1$: The equation sin(x)=a has two solutions per period: x1 and x2=π−x1 (where x1=arcsin(a)). The sign of a determines the quadrants. Within one period [0,2π):
sin(x)>a when a>0: the solution is (x1,x2) — the arc where sine exceeds a, passing through the maximum.
sin(x)>a when a<0: the solution is [0,x2)∪(x1,2π) where x1=π−arcsin(a) and x2=arcsin(a)+2π — the larger arc that includes the maximum. More carefully: the two boundary angles are π+∣arcsin(a)∣ and 2π−∣arcsin(a)∣ (in Quadrants III and IV), and sine exceeds a on the arc that does not pass through the minimum.
Case $a = 1$:sin(x)≥1 is satisfied only at x=2π+2nπ (a single point per period). sin(x)>1 has no solution.
Case $a = -1$:sin(x)≤−1 is satisfied only at x=23π+2nπ. sin(x)<−1 has no solution.
Case $|a| > 1$:sin(x)>a when a>1: no solution (sine never exceeds 1). sin(x)>a when a<−1: all real numbers (sine is always greater than any value below −1).
These cases follow directly from the boundedness of sine. Checking whether a falls within [−1,1] before proceeding prevents unnecessary computation.
Solving Cosine and Tangent Inequalities
Cosine inequalities follow the same logic as sine, with the roles of the coordinates swapped. The equation cos(x)=a (for ∣a∣<1) has solutions x1=arccos(a) and x2=2π−arccos(a) on [0,2π). The key difference from sine: cosine is decreasing on [0,π] and increasing on [π,2π], so the shape of the solution interval is different.
cos(x)>a on [0,2π): the arc where cosine exceeds a is [0,x1)∪(x2,2π) — the region near θ=0 (and 2π), where the x-coordinate on the unit circle is large.
cos(x)<a on [0,2π): the complementary arc (x1,x2).
The boundary cases (a=±1, ∣a∣>1) parallel those for sine.
Tangent inequalities have a different structure because tangent has period π, vertical asymptotes, and unbounded range. The equation tan(x)=a has one solution per period: x0=arctan(a). Within one period (−2π,2π):
tan(x)>a: the interval (x0,2π) — from the boundary angle to the next asymptote.
tan(x)<a: the interval (−2π,x0) — from the asymptote to the boundary angle.
Since tangent is strictly increasing on each period interval, there is no ambiguity: the function is below a to the left of x0 and above a to the right. The general solution adds nπ to shift across periods.
The asymptotes must always be excluded from the solution, regardless of whether the inequality is strict or non-strict — tangent is undefined there.
Compound Inequalities
A compound trigonometric inequality imposes two simultaneous bounds. For example:
21<sin(x)≤23
This requires sin(x)>21 and sin(x)≤23 simultaneously.
Step 1: Solve each inequality separately.
sin(x)>21 on [0,2π): x∈(6π,65π)
sin(x)≤23 on [0,2π): x∈[0,3π]∪[32π,2π)
Step 2: Intersect the solution sets.
The intersection is the set of x values that satisfy both conditions:
x∈(6π,3π]∪[32π,65π)
Graphically, this corresponds to the region where the sine curve lies strictly above y=21 and at or below y=23 — the band between the two horizontal lines, with appropriate endpoint inclusion.
The unit circle interpretation is equally clear: identify the arcs satisfying each condition, then take their overlap. For the example above, the first condition selects the upper arc from 6π to 65π, and the second condition excludes the portion between 3π and 32π (where sine exceeds 23). What remains is two symmetric segments flanking the excluded zone.
Inequalities Involving Transformed Functions
When the trigonometric function is transformed — 2sin(3x)−1>0, for instance — the inequality must be handled in stages.
Step 1: Isolate the trigonometric expression. Rearrange to place the function on one side:
2sin(3x)>1⇒sin(3x)>21
Step 2: Solve for the inner argument. Let u=3x. The inequality sin(u)>21 has the solution:
u∈(6π+2nπ,65π+2nπ)
Step 3: Recover $x$. Divide by 3:
x∈(18π+32nπ,185π+32nπ)
Step 4: Restrict to the target interval. On [0,2π), substitute n=0,1,2 (since the period of sin(3x) is 32π, three periods fit within [0,2π)):
n=0: (18π,185π)
n=1: (18π+32π,185π+32π)=(1813π,1817π)
n=2: (18π+34π,185π+34π)=(1825π,1829π)
The multiplier in the argument multiplies the number of solution intervals within a fixed domain — the same phenomenon seen in multiple-angle equations. A coefficient of B in sin(Bx) creates B copies of the basic solution pattern within one 2π interval.
Vertical shifts (+D) and amplitude changes (A) affect the threshold value after isolation. Phase shifts (−C) offset the interval boundaries. But the core method remains the same: isolate, solve for the inner argument, recover the outer variable, and restrict to the domain.
Domain Considerations for Reciprocal and Quotient Functions
Inequalities involving tangent, cotangent, secant, or cosecant require attention to the domains of these functions — specifically, the vertical asymptotes where they are undefined.
Example: Solve sec(x)≥2 on [0,2π).
Rewrite using the reciprocal: cos(x)1≥2. This requires two conditions simultaneously:
1. cos(x)>0 (secant is positive only when cosine is positive) 2. cos(x)≤21 (taking the reciprocal of both sides reverses the inequality, since cos(x)>0)
The intersection of cos(x)>0 and cos(x)≤21 on [0,2π):
cos(x)>0 on [0,2π)∪(23π,2π)
cos(x)≤21 on [3π,35π]
Intersection: [3π,2π)∪(23π,35π]
The points x=2π and x=23π are excluded because secant is undefined there (vertical asymptotes on the graph).
The reversal of the inequality direction when taking the reciprocal deserves emphasis. If cos(x)>0 and cos(x)1≥2, then cos(x)≤21. If cos(x)<0 and cos(x)1≥2, there is no solution — a negative reciprocal cannot exceed 2. This case analysis — splitting based on the sign of the denominator — is essential for all inequalities involving reciprocal or quotient functions.