The general solution of the equation $(\sqrt 3 - 1)\sin \theta + (\sqrt 3 + 1)\cos \theta = 2$ is
$2n\pi \pm \frac{\pi }{4} + \frac{\pi }{{12}}$
$n\pi + {( - 1)^n}\frac{\pi }{4} + \frac{\pi }{{12}}$
$2n\pi \pm \frac{\pi }{4} - \frac{\pi }{{12}}$
$n\pi + {( - 1)^n}\frac{\pi }{4} - \frac{\pi }{{12}}$
One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval
The solution set of the system of equation
$x\,\, + \,\,y\,\, = \,\,\frac{{2\pi }}{3},\,{\rm{cos}}\,{\rm{x + }}\,{\rm{ cos}}\,{\rm{y}}\,{\rm{ = }}\,\frac{3}{2},$ where $x$ and $y$ are real in
If $r\,\sin \theta = 3,r = 4(1 + \sin \theta ),\,\,0 \le \theta \le 2\pi ,$ then $\theta = $
If $1 + \cot \theta = {\rm{cosec}}\theta $, then the general value of $\theta $ is
Let $A = \left\{ {\theta \,:\,\sin \,\left( \theta \right) = \tan \,\left( \theta \right)} \right\}$ and $B = \left\{ {\theta \,:\,\cos \,\left( \theta \right) = 1} \right\}$ be two sets. Then