The general solution of $\sin x - 3\sin 2x + \sin 3x = $ $\cos x - 3\cos 2x + \cos 3x$ is

Let $f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x$ $+C \cos 2 x+D \cos x+E$, and

$T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) \text {. }$Then, $T$

The number of values of $x$ for which $sin\,\, 2x + cos\,\, 4x = 2$ is

Let $S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$ and $\beta=\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right)$, then $\frac{1}{6}(\beta-14)^2$ is equal to

The most general value of $\theta $ satisfying the equations $\tan \theta = - 1$ and $\cos \theta = \frac{1}{{\sqrt 2 }}$ is

If $(1 + \tan \theta )(1 + \tan \phi ) = 2$, then $\theta + \phi  =$ ....$^o$