5.10 Solving trigonometric equations in an interval — Practise

Work through these on your own. Sketch the curve over the interval first, and write down how many solutions you expect. Give every solution, and give angles in radians.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Solve $\sin\theta = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$ .
Solve $\cos\theta = -\dfrac{1}{2}$ for $0 \le \theta < 2\pi$ .
Solve $\tan\theta = \sqrt{3}$ for $0 \le \theta < 2\pi$ .
Solve $\sin 2\theta = \dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$ .
Solve $2\cos^2\theta - \cos\theta = 0$ for $0 \le \theta < 2\pi$ .
Solve $\sin^2\theta = \dfrac{1}{4}$ for $0 \le \theta < 2\pi$ .
Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0 \le \theta < 2\pi$ .
Solve $\sin\theta = \dfrac{1}{2}$ for $0 \le \theta < 2\pi$ .
Solve $\cos\theta = -\dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$ .
Solve $\tan\theta = 1$ for $0 \le \theta < 2\pi$ .

Answers and concise solutions

1. The solutions

Answer: Sine is positive in the first and second quadrants. The principal value is $\dfrac{\pi}{3}$ , and the second solution is $\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$ .

2. The solutions

Answer: The related acute angle is $\dfrac{\pi}{3}$ . Cosine is negative in the second and third quadrants, so the solutions are $\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$ and $\pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$ .

3. The solutions

Answer: The principal value is $\dfrac{\pi}{3}$ . Tangent has period $\pi$ , so the second solution is $\dfrac{\pi}{3} + \pi = \dfrac{4\pi}{3}$ .

4. The solutions

Answer: Let $u = 2\theta$ , so $0 \le u < 4\pi$ . The equation $\sin u = \dfrac{\sqrt{3}}{2}$ has four solutions in that range: $u = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{7\pi}{3}, \dfrac{8\pi}{3}$ . Dividing by $2$ gives $\theta = \dfrac{\pi}{6}, \dfrac{\pi}{3}, \dfrac{7\pi}{6}, \dfrac{4\pi}{3}$ .

5. The solutions

Answer: Factorise rather than divide: $\cos\theta(2\cos\theta - 1) = 0$ . So $\cos\theta = 0$ , giving $\theta = \dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$ , or $\cos\theta = \dfrac{1}{2}$ , giving $\theta = \dfrac{\pi}{3}$ and $\dfrac{5\pi}{3}$ .

6. The solutions

Answer: Take both square roots: $\sin\theta = \dfrac{1}{2}$ or $\sin\theta = -\dfrac{1}{2}$ . The first gives $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$ ; the second gives $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$ .

7. The solutions

Answer: Factorise: $(2\sin\theta + 1)(\sin\theta - 1) = 0$ . So $\sin\theta = -\dfrac{1}{2}$ or $\sin\theta = 1$ . The first gives $\theta = \dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$ ; the second gives $\theta = \dfrac{\pi}{2}$ . The three solutions are $\dfrac{\pi}{2}$ , $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$ .

8. The solutions

Answer: The principal value is $\dfrac{\pi}{6}$ . The second solution is $\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ .

9. The solutions

Answer: The related acute angle is $\dfrac{\pi}{6}$ . Cosine is negative in the second and third quadrants, so $\theta = \dfrac{5\pi}{6}$ and $\dfrac{7\pi}{6}$ .

10. The solutions

Answer: The principal value is $\dfrac{\pi}{4}$ . Adding the period $\pi$ gives the second solution $\dfrac{5\pi}{4}$ .