The general solution of the equation $sin^{100}x\,-\,cos^{100} x= 1$ is
$2n\pi + \frac{\pi }{3},\,n \in I$
$n\pi + \frac{\pi }{2},\,n \in I$
$n\pi + \frac{\pi }{4},\,n \in I$
$2n\pi - \frac{\pi }{3},\,n \in I$
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
If $2{\cos ^2}x + 3\sin x - 3 = 0,\,\,0 \le x \le {180^o}$, then $x =$
Solve $2 \cos ^{2} x+3 \sin x=0$
One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval
The smallest positive values of $x$ and $y$ which satisfy $\tan (x - y) = 1,\,$ $\sec (x + y) = \frac{2}{{\sqrt 3 }}$ are