Solving Trigonometric Equations

A trig identity like $\sin^2\theta + \cos^2\theta = 1$ is true for every angle. A trig equation like $\sin x = 0.5$ is true only for specific angles. This page teaches you to find those angles — all of them.

The key insight: because trig functions are periodic, most trig equations have infinitely many solutions. Your job is to find the solutions in one period, then use the period to generate the rest.

Solving Simple Equations on the Unit Circle

Example 1: Solve $\sin x = \frac{1}{2}$ in $[0, 2\pi)$

Step 1 — Find the reference angle. Since $\sin\!\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the reference angle is $\frac{\pi}{6}$.

Step 2 — Determine which quadrants. Sine is positive in Quadrants I and II (from the unit circle).

Step 3 — Find all solutions in $[0, 2\pi)$:

• Quadrant I: $x = \frac{\pi}{6}$
• Quadrant II: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Answer: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$

General Solutions: The $+ 2n\pi$ Pattern

If you need all solutions (not just those in one period), add the period to each solution:

$x = \frac{\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2n\pi \quad \text{where } n \text{ is any integer}$

The period of sine and cosine is $2\pi$. For tangent, the period is $\pi$, so tangent equations use $+ n\pi$.

Example 2: Solve $\cos x = -\frac{\sqrt{3}}{2}$, all solutions

Reference angle: $\cos^{-1}\!\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$

Quadrants: Cosine is negative in Quadrants II and III.

Solutions in $[0, 2\pi)$:

• Quadrant II: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
• Quadrant III: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$

General solutions:

$x = \frac{5\pi}{6} + 2n\pi \quad \text{and} \quad x = \frac{7\pi}{6} + 2n\pi$

Example 3: Solve $\tan x = 1$, all solutions

Reference angle: $\tan^{-1}(1) = \frac{\pi}{4}$

Since tangent has period $\pi$ and $\tan\!\left(\frac{\pi}{4}\right) = 1$:

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

This single expression captures all solutions: $\frac{\pi}{4}, \frac{5\pi}{4}, -\frac{3\pi}{4}, \ldots$

Isolating the Trig Function First

When the trig function is not already isolated, solve for it first — just like isolating $x$ in algebra.

Example 4: Solve $2\sin x - 1 = 0$ in $[0, 2\pi)$

$2\sin x = 1$

$\sin x = \frac{1}{2}$

This is the same equation as Example 1. Solutions: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.

Equations That Factor

When a trig equation has multiple terms, factor and apply the zero product property — the same technique used in algebra with solving equations.

Example 5: Solve $2\sin^2 x - \sin x = 0$ in $[0, 2\pi)$

Factor out $\sin x$:

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

Set each factor to zero:

$\sin x = 0 \quad \text{or} \quad 2\sin x - 1 = 0$

$\sin x = 0 \quad \text{or} \quad \sin x = \frac{1}{2}$

Solve each:

• $\sin x = 0$: $x = 0, \pi$
• $\sin x = \frac{1}{2}$: $x = \frac{\pi}{6}, \frac{5\pi}{6}$

Answer: $x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi$

Critical warning: Never divide both sides by $\sin x$ — you would lose the $\sin x = 0$ solutions. Always factor instead.

Common Mistakes

1. Dividing by a trig function instead of factoring. If you divide both sides of $\sin x(2\sin x - 1) = 0$ by $\sin x$, you lose the solutions where $\sin x = 0$. Always factor and use the zero product property.
2. Forgetting the second solution in each period. $\sin x = \frac{1}{2}$ has two solutions in $[0, 2\pi)$, not one. Sine and cosine equations almost always have two solutions per period.
3. Confusing “all solutions” with “solutions in $[0, 2\pi)$.” Read the problem carefully — look for whether it specifies a restricted interval like “in $[0, 2\pi)$” or asks for “all solutions” (which requires the $+ 2n\pi$ terms). If no domain is specified, you must typically provide all solutions.
4. Only using the calculator value. Your calculator gives one value (the principal value from the inverse function). You must use the unit circle to find the other solutions in the period.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• A trig equation is true for specific angles, not all angles — unlike an identity
• For sine and cosine: find the reference angle, determine which quadrants, and locate two solutions per period
• For tangent: the period is $\pi$, so general solutions use $+ n\pi$ (giving one solution per half-period)
• Always factor instead of dividing by a trig function — division loses solutions
• “All solutions” means include the $+ 2n\pi$ (or $+ n\pi$) terms; solutions “in $[0, 2\pi)$” means list only the specific values in that interval

For equations that require identity substitution, quadratic techniques, or multiple-angle methods, continue to Advanced Trigonometric Equations.

Return to Trigonometry for more topics in this section.