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3.10 - Trigonometric Equations and Inequalities

Overview

Now inverse trig functions go to work. In this topic you'll solve equations like 2cos⁡θ=32\cos\theta = \sqrt{3} and inequalities like cos⁡θ<−12\cos\theta < -\frac{1}{2}. An inverse trig function hands you one solution. Because trig functions repeat, there are usually more, and finding all of them in the domain that matters is the real skill. Free-Response Question 4 on the AP exam (Symbolic Manipulations) can include an equation like these.

Inverse trig gives one solution, not all

To solve sin⁡θ=k\sin\theta = k, you can apply arcsine: θ=arcsin⁡k\theta = \arcsin k. But arcsine only returns the angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Other angles have the same sine, and you find them with the symmetry of the unit circle:
Equation
First solution
Other solution in the same cycle
Why
sin⁡θ=k\sin\theta = k
θ1=arcsin⁡k\theta_1 = \arcsin k
π−θ1\pi - \theta_1
θ\theta and π−θ\pi - \theta have the same yy-coordinate on the unit circle
cos⁡θ=k\cos\theta = k
θ1=arccos⁡k\theta_1 = \arccos k
−θ1-\theta_1 (or 2π−θ12\pi - \theta_1)
θ\theta and −θ-\theta have the same xx-coordinate
tan⁡θ=k\tan\theta = k
θ1=arctan⁡k\theta_1 = \arctan k
θ1+π\theta_1 + \pi (next period)
tangent has period π\pi, so there's one solution per period
So the inverse function's answer often needs adjusting: you may need to add the second solution, or shift the answer by a period to land in the interval the problem asks for.

Infinitely many solutions

Sine and cosine repeat every 2π2\pi, so once you have the solutions in one cycle, adding any whole number of periods gives the rest:
θ=θ1+2πkorθ=θ2+2πk,k an integer\theta = \theta_1 + 2\pi k \quad \text{or} \quad \theta = \theta_2 + 2\pi k, \quad k \text{ an integer}
For tangent one family covers everything: θ=θ1+πk\theta = \theta_1 + \pi k.
3.10 - Trigonometric Equations and Inequalities
The line y=12y = \frac{1}{2} crosses y=sin⁡θy = \sin\theta twice in every cycle. In [0,2π)[0, 2\pi) the crossings are at π6\frac{\pi}{6} and 5π6\frac{5\pi}{6}, and every other crossing is one of these plus a multiple of 2π2\pi.

Domain restrictions from context

Most problems don't want infinitely many answers. Either they give an interval, like 0≤θ<2π0 \le \theta < 2\pi, or the context limits the inputs: time can't be negative, a model only covers one day, a rider is on the wheel for one revolution. Find the general solutions, then keep only the ones in the allowed domain.

A method that always works

  1. Isolate the trig function, so the equation looks like sin⁡(input)=k\sin(\text{input}) = k.
  1. Check that a solution can exist. For sine and cosine, kk has to be between −1-1 and 11.
  1. Find the solutions in one cycle: the unit circle for special values, inverse trig plus symmetry for the rest.
  1. Add multiples of the period to get every solution.
  1. Keep only the solutions in the required domain. If the input is bθb\theta, solve for bθb\theta over a stretched interval first, then divide by bb.

Example 1: A basic equation

Solve 2cos⁡θ−3=02\cos\theta - \sqrt{3} = 0 for all θ\theta. Then list the solutions in [0,2π)[0, 2\pi).
Step 1: Isolate: cos⁡θ=32\cos\theta = \frac{\sqrt{3}}{2}.
Step 2: On the unit circle, cosine is 32\frac{\sqrt{3}}{2} at θ=π6\theta = \frac{\pi}{6}, and by symmetry at θ=−π6\theta = -\frac{\pi}{6}, which is the same angle as 11π6\frac{11\pi}{6}.
Step 3: All solutions:
θ=π6+2πkorθ=11π6+2πk,k an integer\theta = \frac{\pi}{6} + 2\pi k \quad \text{or} \quad \theta = \frac{11\pi}{6} + 2\pi k, \quad k \text{ an integer}
Answer: in [0,2π)[0, 2\pi), θ=π6\theta = \frac{\pi}{6} and θ=11π6\theta = \frac{11\pi}{6}.

Example 2: Values that aren't special

Solve on [0,2π)[0, 2\pi): (a) 3sin⁡θ=23\sin\theta = 2 (b) tan⁡θ=−2\tan\theta = -2
(a) sin⁡θ=23\sin\theta = \frac{2}{3}, which isn't a unit circle value, so use arcsine:
θ1=arcsin⁡(23)≈0.730θ2=π−θ1≈2.412\theta_1 = \arcsin\left(\frac{2}{3}\right) \approx 0.730 \qquad \theta_2 = \pi - \theta_1 \approx 2.412
Both are in [0,2π)[0, 2\pi). Sine is positive in Quadrants I and II, which matches.
(b) arctan⁡(−2)≈−1.107\arctan(-2) \approx -1.107, which is not in [0,2π)[0, 2\pi). Tangent has period π\pi, so add π\pi until you land in the interval:
−1.107+π≈2.034−1.107+2π≈5.176-1.107 + \pi \approx 2.034 \qquad -1.107 + 2\pi \approx 5.176
Answer: (a) θ≈0.730\theta \approx 0.730 and θ≈2.412\theta \approx 2.412. (b) θ≈2.034\theta \approx 2.034 and θ≈5.176\theta \approx 5.176.

Example 3: Factoring

Solve 2sin⁡2θ−sin⁡θ−1=02\sin^2\theta - \sin\theta - 1 = 0 on [0,2π)[0, 2\pi).
Step 1: This is a quadratic in sin⁡θ\sin\theta. Let u=sin⁡θu = \sin\theta:
2u2−u−1=0⇒(2u+1)(u−1)=02u^2 - u - 1 = 0 \quad \Rightarrow \quad (2u + 1)(u - 1) = 0
Step 2: So sin⁡θ=−12\sin\theta = -\frac{1}{2} or sin⁡θ=1\sin\theta = 1.
  • sin⁡θ=−12\sin\theta = -\frac{1}{2} at θ=7π6\theta = \frac{7\pi}{6} and θ=11π6\theta = \frac{11\pi}{6} (Quadrants III and IV).
  • sin⁡θ=1\sin\theta = 1 at θ=π2\theta = \frac{\pi}{2}.
Answer: θ=π2,7π6,11π6\theta = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}.

Example 4: An equation with bθb\theta

Solve sin⁡(2θ)=32\sin(2\theta) = \frac{\sqrt{3}}{2} on [0,2π)[0, 2\pi).
Step 1: Let u=2θu = 2\theta. If 0≤θ<2π0 \le \theta < 2\pi, then 0≤u<4π0 \le u < 4\pi: two full cycles.
Step 2: sin⁡u=32\sin u = \frac{\sqrt{3}}{2} at u=π3u = \frac{\pi}{3} and u=2π3u = \frac{2\pi}{3} in the first cycle, and at π3+2π=7π3\frac{\pi}{3} + 2\pi = \frac{7\pi}{3} and 2π3+2π=8π3\frac{2\pi}{3} + 2\pi = \frac{8\pi}{3} in the second.
Step 3: Divide by 2.
θ=π6, π3, 7π6, 4π3\theta = \frac{\pi}{6}, \ \frac{\pi}{3}, \ \frac{7\pi}{6}, \ \frac{4\pi}{3}
Four solutions make sense: sin⁡(2θ)\sin(2\theta) has period π\pi, so it completes two cycles on [0,2π)[0, 2\pi), with two solutions in each.

Solving inequalities

To solve a trig inequality, first solve the matching equation to find the boundary points. Then use the graph or the unit circle to decide which intervals between them work.

Example 5: An inequality

Solve 2cos⁡θ+1<02\cos\theta + 1 < 0 on [0,2π)[0, 2\pi).
Step 1: Isolate: cos⁡θ<−12\cos\theta < -\frac{1}{2}.
Step 2: Boundary points: cos⁡θ=−12\cos\theta = -\frac{1}{2} at θ=2π3\theta = \frac{2\pi}{3} and θ=4π3\theta = \frac{4\pi}{3}.
Step 3: Between them, the graph of cos⁡θ\cos\theta is below y=−12y = -\frac{1}{2} (for example cos⁡π=−1\cos\pi = -1). Outside them it's above.
3.10 - Trigonometric Equations and Inequalities
Answer: 2π3<θ<4π3\frac{2\pi}{3} < \theta < \frac{4\pi}{3}. The endpoints aren't included because the inequality is strict, which is why the graph shows open circles.

Example 6: Ferris wheel (AP-style)

From Topic 3.7, a rider's height on a Ferris wheel is h(t)=−25cos⁡(π4t)+30h(t) = -25\cos\left(\frac{\pi}{4}t\right) + 30 meters, tt minutes after boarding. During the first revolution, 0≤t≤80 \le t \le 8, for how many minutes is the rider more than 42.5 meters above the ground?
Step 1: Set up and isolate. Dividing by −25-25 flips the inequality.
−25cos⁡(π4t)+30>42.5⇒−25cos⁡(π4t)>12.5⇒cos⁡(π4t)<−12-25\cos\left(\frac{\pi}{4}t\right) + 30 > 42.5 \quad \Rightarrow \quad -25\cos\left(\frac{\pi}{4}t\right) > 12.5 \quad \Rightarrow \quad \cos\left(\frac{\pi}{4}t\right) < -\frac{1}{2}
Step 2: Use the context's domain. Let u=π4tu = \frac{\pi}{4}t. For 0≤t≤80 \le t \le 8, uu runs from 00 to 2π2\pi, one cycle.
Step 3: From Example 5, cos⁡u<−12\cos u < -\frac{1}{2} when 2π3<u<4π3\frac{2\pi}{3} < u < \frac{4\pi}{3}.
Step 4: Convert back to tt by multiplying by 4π\frac{4}{\pi}:
2π3⋅4π<t<4π3⋅4π⇒83<t<163\frac{2\pi}{3} \cdot \frac{4}{\pi} < t < \frac{4\pi}{3} \cdot \frac{4}{\pi} \quad \Rightarrow \quad \frac{8}{3} < t < \frac{16}{3}
3.10 - Trigonometric Equations and Inequalities
Answer: the rider is above 42.5 meters from about 2.67 to 5.33 minutes after boarding, for a total of 163−83=83≈2.67\frac{16}{3} - \frac{8}{3} = \frac{8}{3} \approx 2.67 minutes, one third of the ride.

Common mistakes

  • Stopping at the calculator's answer. Arcsine, arccosine, and arctangent give one solution; find the rest with symmetry and the period.
  • Losing solutions when the input is bθb\theta. Stretch the interval for bθb\theta first, then divide by bb.
  • Dividing both sides by a trig function. Turning 2cos⁡2θ=cos⁡θ2\cos^2\theta = \cos\theta into 2cos⁡θ=12\cos\theta = 1 loses θ=π2\theta = \frac{\pi}{2} and 3π2\frac{3\pi}{2}, where cos⁡θ=0\cos\theta = 0. Factor instead: cos⁡θ(2cos⁡θ−1)=0\cos\theta(2\cos\theta - 1) = 0.
  • Forgetting to flip the inequality when you multiply or divide by a negative number.
  • Keeping solutions outside the interval or contextual domain.
  • Hunting for solutions that can't exist. sin⁡θ=1.3\sin\theta = 1.3 has none, because sine is never more than 1.

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