Question

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

Answer 1

General solution of equation :

It is given that:

$\begin{array}{rcl}\sqrt{2}\sin 2x& =& 1\\ & \Rightarrow & \sin 2x=\frac{1}{\sqrt{2}}=\sin \left(\frac{\mathrm{\pi }}{4}\right)\end{array}$

We know,

$$\begin{gathered} \sin x=\sin y \\ \Rightarrow x=n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\mathrm{y} \end{gathered}$$

Therefore,

$$\begin{gathered} 2x=n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4} \\ x=\frac{\mathrm{n\pi }}{2}+{(-1)}^{\mathrm{n}}\frac{\mathrm{\pi }}{8} \end{gathered}$$

Hence the value of $x$ is $x=\frac{\mathrm{n\pi }}{2}+{(-1)}^{\mathrm{n}}\frac{\mathrm{\pi }}{8}$.