Solve $\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)$
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 $n \in Z$
The angles $\alpha, \beta, \gamma$ of a triangle satisfy the equations $2 \sin \alpha+3 \cos \beta=3 \sqrt{2}$ and $3 \sin \beta+2 \cos \alpha=1$. Then, angle $\gamma$ equals
Find the principal and general solutions of the equation $\cot x=-\sqrt{3}$
One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval
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(where $sgn(.)$ denotes signum function) -