Question

sin (2 theta) = cos (3 theta)

Answer 1

Answer on Question #52898 – Math – Trigonometry

$$\sin 2 \theta = \cos 3 \theta$$

Solution

Method 1

Since $\sin 2\theta = \cos \left(\frac{\pi}{2} - 2\theta\right)$, cosine is a $2\pi$-periodic, even function, then we obtain

$$\sin 2 \theta = \cos 3 \theta \Rightarrow \cos \left(\frac {\pi}{2} - 2 \theta\right) = \cos 3 \theta \Rightarrow \left[ \begin{array}{l} \frac {\pi}{2} - 2 \theta = 3 \theta - 2 \pi k, k \in Z \\ \frac {\pi}{2} - 2 \theta = - 3 \theta + 2 \pi k, k \in Z \end{array} \right. \Rightarrow \left[ \begin{array}{l} \theta = \frac {1 + 4 k}{10} \pi , k \in Z \\ \theta = \left(2 k - \frac {1}{2}\right) \pi , k \in Z \end{array} \right.$$

Thus we found two collections of solutions

**Answer:** $\begin{bmatrix} \theta = \frac{1 + 4k}{10}\pi ,k\in Z\\ \theta = \left(2k - \frac{1}{2}\right)\pi ,k\in Z \end{bmatrix}$

Method 2

Given $\sin 2\theta = \cos 3\theta$

Taking into account formula $\sin 2\theta = \cos \left(\frac{\pi}{2} - 2\theta\right)$, obtain that

$$\cos \left(\frac {\pi}{2} - 2 \theta\right) = \cos 3 \theta$$

Apply the formula $\cos x - \cos y = -2\sin \left(\frac{x + y}{2}\right)\sin \left(\frac{x - y}{2}\right)$ to the previous equality and come to

$$\begin{array}{l} \cos \left(\frac {\pi}{2} - 2 \theta\right) - \cos 3 \theta = 0 \\ - 2 \sin \left(\frac {\pi / 2 - 2 \theta + 3 \theta}{2}\right) \sin \left(\frac {\pi / 2 - 2 \theta - 3 \theta}{2}\right) = 0 \\ \sin \left(\frac {\pi / 2 - 2 \theta + 3 \theta}{2}\right) = 0 \quad \text{or} \quad \sin \left(\frac {\pi / 2 - 2 \theta - 3 \theta}{2}\right) = 0 \\ \end{array}$$

Recall that $\sin (x) = 0$ gives $x = k\pi$, $k$ is integer. Thus,

$$\begin{array}{l} \frac {\pi}{4} + \frac {\theta}{2} = k \pi , \text{ that is,} \quad \theta = \left(2 k - \frac {1}{2}\right) \pi , \text{ or} \\ \frac {\pi}{4} - \frac {10 \theta}{4} = k \pi , \text{ denote } k = -l, \text{ therefore } \frac {\pi}{4} - \frac {10 \theta}{4} = -l \pi , \text{ hence } \theta = \frac {4l + 1}{10}. \\ \end{array}$$

**Answer:** $\begin{bmatrix} \theta = \frac{1 + 4k}{10}\pi ,k\in Z\\ \theta = \left(2k - \frac{1}{2}\right)\pi ,k\in Z \end{bmatrix}$

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