Solving Trig Equations

The recipe

To solve an equation like $2 \sin \theta = 1$ on $[0, 2\pi)$:

1. Isolate the trig function.
2. Find the reference solution from the unit circle.
3. Use symmetry to find all solutions in $[0, 2\pi)$.
4. (If asked) add multiples of the period for the general solution.

Worked example 1

Solve $2 \sin \theta = 1$ on $[0, 2\pi)$.

1. Isolate: $\sin \theta = \dfrac{1}{2}$.
2. Reference solution: $\theta = \pi/6$.
3. Sine is positive in quadrants I and II. The second solution is $\pi - \pi/6 = 5\pi/6$.

Solutions: $\theta = \pi/6,\ 5\pi/6$.

Worked example 2 — factor first

Solve $\sin^2 \theta - \sin \theta = 0$ on $[0, 2\pi)$.

Factor: $\sin \theta (\sin \theta - 1) = 0$.

So $\sin \theta = 0$ or $\sin \theta = 1$.

• $\sin \theta = 0 \implies \theta = 0,\ \pi$.
• $\sin \theta = 1 \implies \theta = \pi/2$.

Solutions: $\theta = 0,\ \pi/2,\ \pi$.

Worked example 3 — Pythagorean substitution

Solve $2 \sin^2 \theta + 3 \cos \theta = 0$ on $[0, 2\pi)$.

Convert sine to cosine using $\sin^2 = 1 - \cos^2$:

$2(1 - \cos^2 \theta) + 3 \cos \theta = 0$ $-2 \cos^2 \theta + 3 \cos \theta + 2 = 0$ $2 \cos^2 \theta - 3 \cos \theta - 2 = 0$

Let $u = \cos \theta$: $2u^2 - 3u - 2 = 0 \implies (2u + 1)(u - 2) = 0$.

So $u = -\dfrac{1}{2}$ or $u = 2$. The second is impossible (cosine $\leq 1$). Take $\cos \theta = -\dfrac{1}{2}$.

Cosine is negative in Q2 and Q3. Reference angle: $\pi/3$. Solutions: $\pi - \pi/3 = 2\pi/3$ and $\pi + \pi/3 = 4\pi/3$.

Solutions: $\theta = 2\pi/3,\ 4\pi/3$.

General solutions

To capture all real-number solutions, add multiples of the period. For sine and cosine, add $2\pi k$. For tangent, add $\pi k$.

Example. All solutions of $\sin \theta = \dfrac{1}{2}$:

$\theta = \dfrac{\pi}{6} + 2\pi k \quad \text{or} \quad \theta = \dfrac{5\pi}{6} + 2\pi k$

for any integer $k$.

Key takeaways

• Isolate the trig function; use the unit circle for the reference solution.
• Look for factoring or a Pythagorean substitution to reduce to one trig function.
• For "all solutions," add $2\pi k$ (sine/cosine) or $\pi k$ (tangent).