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