Question

Solve the equation on the interval $0\le \theta \le 2\mathrm{\pi }$.

$\sin \left(2\theta \right)=\sqrt{2}\cos \theta$

Short answer

On solving the equation, we get,

$\theta \in \left\{\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{2}\right\}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} \sin \left(2\theta \right)=\sqrt{2}\cos \theta \\ 0\le \theta \le 2\mathrm{\pi } \end{gathered}$$

We know $\sin \left(2\theta \right)=2\sin \theta ·\cos \theta$,

$$\begin{gathered} 2\sin \theta ·\cos \theta =\sqrt{2}\cos \theta \\ 2\sin \theta ·\cos \theta -\sqrt{2}\cos \theta =0 \\ c\mathrm{os}\theta \left(2\sin \theta -\sqrt{2}\right)=0 \end{gathered}$$

Then,

localid="1646750340707" $\cos \theta =0$

Also,

$$\begin{gathered} 2\sin \theta -\sqrt{2}=0 \\ \sin \theta =\frac{\sqrt{2}}{2} \\ \sin \theta =\frac{1}{\sqrt{2}} \end{gathered}$$

Step 2. Solve the equation for the given interval.

Consider the given question,

$$\begin{gathered} \cos \theta =0 \\ \sin \theta =\frac{1}{\sqrt{2}} \\ 0\le \theta \le 2\mathrm{\pi } \end{gathered}$$

Solving $\cos \theta =0$for the given interval,

$$\begin{gathered} \cos \theta =0 \\ \theta ={\cos }^{-1}\left(0\right) \\ \theta =\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2} \end{gathered}$$

Solving role="math" localid="1646750410065" $\sin \theta =\frac{1}{\sqrt{2}}$for the given interval,

$$\begin{gathered} \sin \theta =\frac{1}{\sqrt{2}} \\ \theta ={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ \theta =\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4} \end{gathered}$$

Therefore, the solution set is$\theta \in \left\{\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{2}\right\}$.