Let $f:[0,2] \rightarrow R$ be the function defined by
$f ( x )=(3-\sin (2 \pi x )) \sin \left(\pi x -\frac{\pi}{4}\right)-\sin \left(3 \pi x +\frac{\pi}{4}\right)$
If $\alpha, \beta \in[0,2]$ are such that $\{x \in[0,2]: f(x) \geq 0\}=[\alpha, \beta]$, then the value of $\beta-\alpha$ is. . . . . . . . .
$0$
$1$
$5$
$6$
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
The number of solution of the equation,$\sum\limits_{r = 1}^5 {\cos (r\,x)} $ $= 0$ lying in $(0, \pi)$ is :
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 :-
If $\sin \,\theta + \sqrt 3 \cos \,\theta = 6x - {x^2} - 11,x \in R$ , $0 \le \theta \le 2\pi $ , then the equation has solution for
If ${\left( {\frac{{\sin \theta }}{{\sin \phi }}} \right)^2} = \frac{{\tan \theta }}{{\tan \phi }} = 3,$ then the value of $\theta $ and $\phi $ are