Question

How do you solve $2\sin 3x=\sqrt{2}$ on the interval $[0,2\mathrm{\pi })$?

Answer 1

Solving the given trigonometric equation.

Given the trigonometric equation: $2\sin 3x=\sqrt{2}$

Dividing both sides by 2:

$\begin{array}{rcl}\sin 3x& =& \frac{\sqrt{2}}{2}\\ & \Rightarrow & \sin 3x=\frac{1}{\sqrt{2}}\\ & \Rightarrow & 3x={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)\end{array}$

The angle $3x$ can be $\frac{\mathrm{\pi }}{4}$ or $\frac{3\mathrm{\pi }}{4}$ in the interval $[0,2\mathrm{\pi })$.

Hence, $x$ can be $\frac{\mathrm{\pi }}{12}$or $\frac{\mathrm{\pi }}{4}$in the interval $[0,2\mathrm{\pi })$.

Therefore, the solution set of the given trigonometric equation, $\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{3}\mathit{x}\mathbf{=}\sqrt{\mathbf{2}}$, in the interval $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{\pi }\mathbf{)}$ is $\left\{\frac{\pi }{12},\frac{\pi }{4}\right\}$.