Advanced Trigonometric Equations

The basic equation-solving techniques from Solving Trigonometric Equations work when you can isolate a single trig function directly. But many equations involve two different trig functions, or powers of trig functions, or multiple angles. This page covers three techniques that handle these harder cases.

Technique 1: Use an Identity to Reduce to One Function

When an equation has two different trig functions (like both sine and cosine), use a trig identity to rewrite everything in terms of a single function.

Example 1: Solve $2\sin^2 x + \cos x = 1$ in $[0, 2\pi)$

This has both $\sin^2 x$ and $\cos x$. Use the Pythagorean identity $\sin^2 x = 1 - \cos^2 x$ to convert to all cosine:

$2(1 - \cos^2 x) + \cos x = 1$

$2 - 2\cos^2 x + \cos x = 1$

$-2\cos^2 x + \cos x + 1 = 0$

Multiply by $-1$:

$2\cos^2 x - \cos x - 1 = 0$

This is a quadratic in $\cos x$.

Technique 2: Quadratic Substitution

Once you have a quadratic in a trig function, let $u = \cos x$ (or $\sin x$) and solve the quadratic — then back-substitute.

Continuing Example 1:

$2u^2 - u - 1 = 0 \quad \text{where } u = \cos x$

Factor:

$(2u + 1)(u - 1) = 0$

$u = -\frac{1}{2} \quad \text{or} \quad u = 1$

Back-substitute:

$\cos x = -\frac{1}{2} \quad \text{or} \quad \cos x = 1$

Solve each:

• $\cos x = -\frac{1}{2}$: $x = \frac{2\pi}{3}, \frac{4\pi}{3}$
• $\cos x = 1$: $x = 0$

Answer: $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$

Important check: When using the quadratic formula (instead of factoring), verify that solutions for $u$ satisfy $-1 \le u \le 1$ for sine and cosine. A value of $u = 2.5$ has no real solutions because no angle has $\sin x = 2.5$.

Example 2: Solve $2\cos^2 x - 3\cos x + 1 = 0$ in $[0, 2\pi)$

Let $u = \cos x$:

$2u^2 - 3u + 1 = 0$

$(2u - 1)(u - 1) = 0$

$u = \frac{1}{2} \quad \text{or} \quad u = 1$

Back-substitute:

• $\cos x = \frac{1}{2}$: $x = \frac{\pi}{3}, \frac{5\pi}{3}$
• $\cos x = 1$: $x = 0$

Answer: $x = 0, \frac{\pi}{3}, \frac{5\pi}{3}$

Technique 3: Multiple-Angle Equations

When the equation involves $\sin(2x)$, $\cos(3x)$, or similar, solve for the inner expression first, then divide.

Example 3: Solve $\sin(2x) = \frac{\sqrt{3}}{2}$ in $[0, 2\pi)$

Step 1 — Let $u = 2x$. Now solve $\sin u = \frac{\sqrt{3}}{2}$.

Step 2 — Adjust the interval. Since $x \in [0, 2\pi)$, we have $u = 2x \in [0, 4\pi)$. We need to search two full periods of sine.

Step 3 — Solve for $u$:

In $[0, 2\pi)$: $u = \frac{\pi}{3}, \frac{2\pi}{3}$

In $[2\pi, 4\pi)$: $u = 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}$ and $u = 2\pi + \frac{2\pi}{3} = \frac{8\pi}{3}$

Step 4 — Divide by 2 to get $x$:

$x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$

Answer: $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$

Key insight: An equation with $\sin(2x)$ has twice as many solutions in $[0, 2\pi)$ as the equivalent $\sin(x)$ equation, because the inner function cycles twice as fast. Similarly, $\cos(3x)$ has three times as many solutions.

Example 4: Solve $\cos(3x) = -\frac{1}{2}$ in $[0, 2\pi)$

Let $u = 3x$, so $u \in [0, 6\pi)$. We need solutions across three full periods.

$\cos u = -\frac{1}{2}$: reference angle $= \frac{\pi}{3}$, cosine negative in Q II and Q III.

In each period $[0, 2\pi)$: $u = \frac{2\pi}{3}, \frac{4\pi}{3}$

All solutions in $[0, 6\pi)$: $u = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{2\pi}{3} + 2\pi, \frac{4\pi}{3} + 2\pi, \frac{2\pi}{3} + 4\pi, \frac{4\pi}{3} + 4\pi$ $= \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}, \frac{14\pi}{3}, \frac{16\pi}{3}$

Divide by 3: $x = \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}, \frac{10\pi}{9}, \frac{14\pi}{9}, \frac{16\pi}{9}$

Answer: $x = \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}, \frac{10\pi}{9}, \frac{14\pi}{9}, \frac{16\pi}{9}$

Strategy Decision Table

Common Mistakes

1. Forgetting to check that quadratic solutions are valid. If $u = \cos x$ and the quadratic gives $u = 1.5$, discard it — no angle has a cosine of 1.5.
2. Not expanding the interval for multiple-angle equations. If $x \in [0, 2\pi)$ and you substitute $u = 2x$, search $u \in [0, 4\pi)$ — not just $[0, 2\pi)$.
3. Introducing extraneous solutions. When you square both sides or use identities, always verify your answers in the original equation.
4. Mixing up $\sin^2 x + \cos x = 1$ with the identity $\sin^2 x + \cos^2 x = 1$. The first is an equation to solve. The second is an identity true for all $x$. They look similar but are fundamentally different.

Practice Problems

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

Key Takeaways

• When an equation has two different trig functions, use a Pythagorean identity to convert to one function
• Quadratic substitution ($u = \sin x$ or $u = \cos x$) turns trig equations into algebra problems — but check that solutions satisfy $|u| \le 1$
• For multiple-angle equations ($\sin(2x)$, $\cos(3x)$), substitute $u = nx$, expand the search interval, and divide at the end
• A $\sin(nx)$ equation has $n$ times as many solutions per period as a $\sin(x)$ equation
• Always verify solutions in the original equation — identity substitutions and squaring can introduce extraneous solutions

Return to Trigonometry for more topics in this section.