7.4 Trigonometric Equations

7.4 Trigonometric Equations

The official content outline lists trigonometric equations inside Algebraic Expressions, Equations, and Inequalities (20%), alongside linear, quadratic, polynomial, rational, radical, exponential, and logarithmic equations. That placement is the point: a trigonometric equation is solved with ordinary algebra, then finished with trigonometric reasoning about how many angles produce the value you found. The second half is where points are lost.

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1. What Makes a Trigonometric Equation Different

A conditional trigonometric equation is true only for particular angles — unlike an identity, which holds for all of them. Because the trigonometric functions are periodic, a conditional equation almost always has infinitely many solutions, and the question will tell you which ones it wants:

Period reminders: $\sin$, $\cos$, $\csc$, and $\sec$ repeat every $2\pi$; $\tan$ and $\cot$ repeat every $\pi$. That difference determines whether you append $2k\pi$ or $k\pi$.

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2. The Four-Step Method

1. Isolate the trigonometric expression, treating it as a single variable.
2. Reduce to one function using identities if two or more different functions appear.
3. Find the reference angle from the exact-value table, then place solutions in every quadrant where the function carries the required sign.
4. Apply the interval — and if the argument is compound (like $2x$ or $x - \frac{\pi}{3}$), solve for the argument over its expanded interval before dividing.

Sign placement, quadrant by quadrant

The mnemonic ASTC ("All Students Take Calculus") names the quadrant in which each function is positive: All in I, Sine in II, Tangent in III, Cosine in IV.

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3. Worked Examples

Worked Example 7.4A: Linear in one function

Solve $2\cos x + \sqrt{3} = 0$ on $[0, 2\pi)$.

Solution:

1. Isolate: $\cos x = -\frac{\sqrt{3}}{2}$.
2. Reference angle: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.
3. Cosine is negative in Quadrants II and III:
   • Quadrant II: $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$
   • Quadrant III: $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$
4. Solution set: $\left{\frac{5\pi}{6}, \frac{7\pi}{6}\right}$.

Worked Example 7.4B: Quadratic form — factor, never divide

Solve $2\sin^2 x - 3\sin x + 1 = 0$ on $[0, 2\pi)$.

Solution:

1. Substitute $u = \sin x$: $2u^2 - 3u + 1 = 0$.
2. Factor: $(2u - 1)(u - 1) = 0 \implies (2\sin x - 1)(\sin x - 1) = 0$.
3. Solve each factor:
   • $\sin x = \frac{1}{2}$: sine is positive in Quadrants I and II, giving $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
   • $\sin x = 1$: this occurs at the single point $x = \frac{\pi}{2}$.
4. Solution set: $\left{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\right}$.

Worked Example 7.4C: Two functions — reduce with an identity

Solve $\cos(2x) + \cos x = 0$ on $[0, 2\pi)$.

Solution:

1. Rewrite the double angle in terms of cosine alone: $\cos(2x) = 2\cos^2 x - 1$.
2. Substitute: $2\cos^2 x - 1 + \cos x = 0$, i.e. $2\cos^2 x + \cos x - 1 = 0$.
3. Factor: $(2\cos x - 1)(\cos x + 1) = 0$.
4. Solve:
   • $\cos x = \frac{1}{2}$: Quadrants I and IV, giving $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.
   • $\cos x = -1$: a single point, $x = \pi$.
5. Solution set: $\left{\frac{\pi}{3}, \pi, \frac{5\pi}{3}\right}$.

Worked Example 7.4D: Compound argument — expand the interval first

Solve $\sin(2x) = \frac{\sqrt{2}}{2}$ on $[0, 2\pi)$.

Solution:

1. Let $u = 2x$. As $x$ runs over $[0, 2\pi)$, the substitution $u$ runs over $[0, 4\pi)$ — two revolutions.
2. Solve $\sin u = \frac{\sqrt{2}}{2}$ over $[0, 4\pi)$. Sine is positive in Quadrants I and II, and the reference angle is $\frac{\pi}{4}$: $u = \frac{\pi}{4}, \; \frac{3\pi}{4}, \; \frac{\pi}{4} + 2\pi = \frac{9\pi}{4}, \; \frac{3\pi}{4} + 2\pi = \frac{11\pi}{4}$
3. Divide each by $2$: $x = \frac{\pi}{8}, \; \frac{3\pi}{8}, \; \frac{9\pi}{8}, \; \frac{11\pi}{8}$

Four solutions, not two. The number of solutions multiplies by the coefficient of $x$.

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4. General Solutions

When a question asks for all solutions, express the answer with an integer parameter $k$:

$\sin x = \frac{1}{2} \implies x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi$

$\tan x = 1 \implies x = \frac{\pi}{4} + k\pi$

Tangent needs only one family because its period is $\pi$ — its Quadrant I and Quadrant III solutions are already $\pi$ apart.

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5. CLEP Exam Traps & Common Errors

• Trap 1: Dividing by a trigonometric factor. In $\sin x\cos x = \sin x$, dividing both sides by $\sin x$ silently destroys the solutions where $\sin x = 0$. Move everything to one side and factor: $\sin x(\cos x - 1) = 0$ gives $x = 0, \pi$ and $x = 0$ from the second factor, so the full set on $[0, 2\pi)$ is ${0, \pi}$.
• Trap 2: Forgetting the extra revolutions with a compound argument. Solving $\cos(3x) = 0$ on $[0, 2\pi)$ requires $3x$ over $[0, 6\pi)$ — six solutions, not two.
• Trap 3: Reporting a solution outside the requested interval. $[0, 2\pi)$ excludes $2\pi$ itself. If your work produces $x = 2\pi$, discard it.
• Trap 4: Missing the second quadrant. After finding a reference angle with an inverse function, you must still place solutions in every quadrant with the correct sign. The calculator returns only one.
• Trap 5: Accepting an out-of-range value. $\sin x = 1.4$ and $\cos x = -3$ have no solutions, because sine and cosine are bounded on $[-1, 1]$. Check the bound before you start solving.