Grade 12 · MHF4U · Advanced Functions

MHF4U: Solving Trig Equations

Solving a trig equation means finding every angle in the interval that works. There is almost always more than one, and the marks are in finding all of them.

How It Works

The method from the lesson, every time:

1. Isolate the trig ratio. Get it to $\sin x =$ a number, the way you would isolate $x$ in any equation.

2. Find the reference angle. Take the inverse of the positive value. For $\sin x = -0.25$ , the reference angle is $\sin^{-1}(0.25)$ . Keep the negative sign out of it.

3. Pick the quadrants with CAST. Sine is positive in quadrants 1 and 2, cosine in 1 and 4, tangent in 1 and 3. A negative value puts you in the other two.

4. Build the angles. In radians: quadrant 1 is $\theta$ , quadrant 2 is $\pi - \theta$ , quadrant 3 is $\pi + \theta$ , quadrant 4 is $2\pi - \theta$ .

5. Check the interval, and for a general solution add $2n\pi$ for sine and cosine, or $n\pi$ for tangent, where $n$ is any integer.

Two upgrades show up on tests. For $\cos 2x = \frac{1}{2}$ , let $\theta = 2x$ . If $0 \le x \le 2\pi$ , then $0 \le \theta \le 4\pi$ , so you go around twice, find four values of $\theta$ , and halve them. Forgetting to widen the interval is the classic lost mark.

The other is a quadratic in disguise. $2\cos^2 x + \cos x - 1 = 0$ is $2c^2 + c - 1 = 0$ with $c = \cos x$ . Factor, solve each factor, and throw out impossible values: $\sin x$ and $\cos x$ never leave $[-1, 1]$ . If the equation mixes $\sin^2 x$ and $\cos x$ , use $\sin^2 x = 1 - \cos^2 x$ first so only one ratio is left.

Exact values come from the special triangles. Anything else, give decimals in radians to four places.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Solve $2\sin x - 1 = 0$ for $0 \le x \le 2\pi$ .
Show solution
1. Isolate: $\sin x = \frac{1}{2}$ .
2. Reference angle: $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ .
3. Sine is positive in quadrants 1 and 2: $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ .
Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}$
Easy
Solve $\sqrt{3}\tan x + 1 = 0$ for $0 \le x \le 2\pi$ .
Show solution
1. Isolate: $\tan x = -\frac{1}{\sqrt{3}}$ .
2. Reference angle from the positive value: $\tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$ .
3. Tangent is negative in quadrants 2 and 4: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ and $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ .
Answer: $x = \frac{5\pi}{6}, \frac{11\pi}{6}$
Medium
Solve $3\cos x + 2 = 0$ for $0 \le x \le 2\pi$ . Give answers to four decimal places.
Show solution
1. Isolate: $\cos x = -\frac{2}{3}$ . Not a special-triangle value, so expect decimals.
2. Reference angle: $\cos^{-1}(\frac{2}{3}) \approx 0.8411$ .
3. Cosine is negative in quadrants 2 and 3.
4. Quadrant 2: $x = \pi - 0.8411 \approx 2.3005$ . Quadrant 3: $x = \pi + 0.8411 \approx 3.9827$ .
Answer: $x \approx 2.3005, 3.9827$
Medium
Solve $2\cos 2x = 1$ for $0 \le x \le 2\pi$ .
Show solution
1. Isolate: $\cos 2x = \frac{1}{2}$ . Let $\theta = 2x$ . Since $0 \le x \le 2\pi$ , $0 \le \theta \le 4\pi$ .
2. Reference angle $\frac{\pi}{3}$ . Cosine is positive in quadrants 1 and 4: $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$ in the first turn.
3. Add $2\pi$ for the second turn: $\theta = \frac{7\pi}{3}, \frac{11\pi}{3}$ .
4. Halve each: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ .
Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$
Test level
Solve $2\sin^2 x = 1 + \cos x$ for $0 \le x \le 2\pi$ .
Show solution
1. Two different ratios, so replace $\sin^2 x$ with $1 - \cos^2 x$ : $2(1 - \cos^2 x) = 1 + \cos x$ .
2. Expand and move everything to one side: $2\cos^2 x + \cos x - 1 = 0$ .
3. Factor as a quadratic in $\cos x$ : $(2\cos x - 1)(\cos x + 1) = 0$ .
4. $\cos x = \frac{1}{2}$ gives $x = \frac{\pi}{3}, \frac{5\pi}{3}$ .
5. $\cos x = -1$ gives $x = \pi$ .
Answer: $x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}$

Learn It on Martian Lab

Martian Lab teaches this in the Trigonometric Equations lesson of its MHF4U Trigonometric Functions unit. Watch it, then take the Trigonometric Functions practice tests to see if it stuck.

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Free Trigonometric Functions practice test, with solutions

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Proving Trig Identities · Laws of Logarithms · Solving Logarithmic Equations

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