Question

Solve the following trigonometric equation:

$\tan 2x=\sqrt{3}$

Answer 1

Trigonometric equations:

We have,

$$\begin{gathered} \tan 2x=\sqrt{3} \\ \Rightarrow \tan 2x=\tan \frac{\mathrm{\pi }}{3}\mathbf{}\mathbf{}\left(\because \tan \frac{\pi }{3}=\sqrt{3}\right) \end{gathered}$$

Now, when we have

$\tan x=\tan y$

then the general value of $x$ is given by

$\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{n\pi }\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{,}\mathbf{}\mathbf{}\mathbf{where}\mathbf{}\mathbf{n}\mathbf{}\mathbf{\in }\mathbf{}\mathbf{Z}$

Therefore,

$\begin{array}{rcl}2x& =& n\mathrm{\pi }+\frac{\mathrm{\pi }}{3},\mathrm{where}\mathrm{n}\in \mathrm{Z}\\ & \Rightarrow & \mathrm{x}=\frac{\mathrm{n\pi }}{2}+\frac{\mathrm{\pi }}{3},\mathrm{where}\mathrm{n}\in \mathrm{Z}\end{array}$

Hence, $\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{\pi }}{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{,}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}\mathbf{n}\mathbf{}\mathbf{\in }\mathbf{}\mathbf{Z}$ is the solution of given trigonometric function.