Find the principal and general solutions of the question $\tan x=\sqrt{3}$.
$\tan x=\sqrt{3}$
It is known that $\tan \frac{\pi}{3}=\sqrt{3}$ and $\tan \left(\frac{4 \pi}{3}\right)=\tan \left(\pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}=\sqrt{3}$
Therefore, the principal solutions are $x=\frac{\pi}{3}$ and $\frac{4 \pi}{3}$
Now, $\tan x=\tan \frac{\pi}{3}$
$\Rightarrow x=n \pi+\frac{\pi}{3},$ where $n \in Z$
Therefore, the general solution is $x=n \pi+\frac{\pi}{3},$ where $n \in Z.$
If $\cos ec\,\theta = \frac{{p + q}}{{p - q}}$ $\left( {p \ne q \ne 0} \right)$, then $\left| {\cot \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right)} \right|$ is equal to
If $tan(\pi sin \theta)$ $= cot(\pi cos \theta)$, then $\left| {\cot \left( {\theta - \frac{\pi }{4}} \right)} \right|$ is -
The value of the expression
$\frac{{\left (sin 36^o + cos 36^o - \sqrt 2 sin 27^o)( {\sin {{36}^0} + \cos {{36}^0} - \sqrt 2 \sin {{27}^0}} \right)}}{{2\sin {{54}^0}}}$ is less than
If $2{\tan ^2}\theta = {\sec ^2}\theta ,$ then the general value of $\theta $ is
The sum of all $x \in[0, \pi]$ which satisfy the equation $\sin x+\frac{1}{2} \cos x=\sin ^2\left(x+\frac{\pi}{4}\right)$ is