Question

How do you solve $2{\sin }^{2}x=1$?

Answer 1

Solving a trigonometric equation:

Given a trigonometric equation,

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

Here, $x$ can be all those angles that are odd multiples of $\frac{\mathrm{\pi }}{4}$.

That is, $\begin{array}{rcl}x& =& \frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}...\frac{\left(2n+1\right)\mathrm{\pi }}{4}\end{array}$ where $n$ is an integer.

Here, $\left(2n+1\right)$ represents an odd number for integer values of $n$.

Therefore, the solution set of the given trigonometric equation is: $\mathit{S}\mathbf{=}\mathbf{\left\{}\mathbf{x}\mathbf{|}\mathbf{}\mathbf{x}\mathbf{=}\frac{\left(2n+1\right)\mathbf{\pi }}{\mathbf{4}}\mathbf{,}\mathbf{}\mathbf{where}\mathbf{}\mathbf{n}\mathbf{\in }\mathbf{Z}\mathbf{\right\}}$