The general solution of $a\cos x + b\sin x = c,$ where $a,\,\,b,\,\,c$ are constants
$x = n\pi + {\cos ^{ - 1}}\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }}} \right)$
$x = 2n\pi - {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)$
$x = 2n\pi - {\tan ^{ - 1}}\left( {\frac{b}{a}} \right) \pm {\cos ^{ - 1}}\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }}} \right)$
$x = 2n\pi + {\tan ^{ - 1}}\left( {\frac{b}{a}} \right) \pm {\cos ^{ - 1}}\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }}} \right)$
If $a = \sin \frac{\pi }{{18}}\sin \frac{{5\pi }}{{18}}\sin \frac{{7\pi }}{{18}}$ and $x$ is the solution of the equatioin $y = 2\left[ x \right] + 2$ and $y = 3\left[ {x - 2} \right] ,$ where $\left[ x \right]$ denotes the integral part of $x,$ then $a$ is equal to :-
Let,$S=\left\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\right\}$. Then $n ( S )+\sum_{\theta \in S}\left(\sec \left(\frac{\pi}{4}+2 \theta\right) \operatorname{cosec}\left(\frac{\pi}{4}+2 \theta\right)\right)$ is equal to.
If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and $M=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right),$ then
If $\sqrt 3 \cos \,\theta + \sin \theta = \sqrt 2 ,$ then the most general value of $\theta $ is
The sum of all values of $x$ in $[0,2 \pi]$, for which $\sin x+\sin 2 x+\sin 3 x+\sin 4 x=0$, is equal to: