Question

Find the principal solution or solution of $\sin x=\frac{1}{\sqrt{2}}$.

Answer 1

Step 1: Simplify using trigonometric standard angles table

It is given that, $\sin x=\frac{1}{\sqrt{2}}$ $\dots \left(\mathrm{i}\right)$

$\Rightarrow$ $\sin x=\sin \frac{\mathrm{\pi }}{4}$ $\dots \left(\mathrm{ii}\right)$ $\left[\because \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right]$

$\Rightarrow$ $\sin x=\sin \left(\pi -\frac{\mathrm{\pi }}{4}\right)$ $[\because \sin \left(\pi -x\right)=\sin x]$

$\Rightarrow$ $\sin x=\sin \left(\frac{4\mathrm{\pi }-\mathrm{\pi }}{4}\right)$

$\Rightarrow$ $\sin x=\sin \frac{3\mathrm{\pi }}{4}$ $\dots \left(\mathrm{iii}\right)$

Step 2: Substitute the value of $\sin x$

From equation $\left(\mathrm{ii}\right)$ in equation $\left(\mathrm{iii}\right)$,

$\Rightarrow$ $\sin \frac{\mathrm{\pi }}{4}=\sin \frac{3\mathrm{\pi }}{4}$

$\Rightarrow$ $\sin \frac{\mathrm{\pi }}{4}=\sin \frac{3\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}$ $\dots \left(\mathrm{iv}\right)$ $[\because \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}},\sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}]$

Where, $0<\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}<2\mathrm{\pi }$

From equation $\left(\mathrm{i}\right)$ and equation $\left(\mathrm{iv}\right)$,

$\sin x=\sin \frac{\mathrm{\pi }}{4}=\sin \frac{3\mathrm{\pi }}{4}$

So, $x=\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}$

Hence, the required solution is $x=\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}$.