Solve $\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

We have, $\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)$

$=\tan \left(\frac{\pi}{2}+x+\frac{\pi}{3}\right)$

or $\tan 2 x=\tan \left(x+\frac{5 \pi}{6}\right)$

Therefore $2 x=n \pi+x+\frac{5 \pi}{6},$ where $n \in Z$

or $x=n \pi+\frac{5 \pi}{6},$ where $n \in Z$

Similar Questions

$2{\sin ^2}x + {\sin ^2}2x = 2,\, - \pi < x < \pi ,$ then $x = $

If $\operatorname{cosec}^2(\alpha+\beta)-\sin ^2(\beta-\alpha)+\sin ^2(2 \alpha-\beta)=\cos ^2(\alpha-\beta)$ where $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, then $\sin (\alpha-\beta)$ is equal to

  • [KVPY 2009]

If the solution for $\theta $ of $\cos p\theta + \cos q\theta = 0,\;p > 0,\;q > 0$ are in $A.P.$, then the numerically smallest common difference of $A.P.$ is

The equation $\sin x\cos x = 2$ has

The equation $2{\cos ^2}\left( {\frac{x}{2}} \right)\,{\sin ^2}x\, = \,{x^2}\, + \,\frac{1}{{{x^2}}},\,0\,\, \leqslant \,\,x\,\, \leqslant \,\,\frac{\pi }{2}\,\,$ has