If $x = \frac{{n\pi }}{2}$ , satisfies the equation $sin\, \frac{x}{2}- cos \frac{x}{2} = 1$ $- sin\, x$ & the inequality $\left| {\frac{x}{2}\,\, - \,\,\frac{\pi }{2}} \right|\,\, \le \,\,\frac{{3\pi }}{4}$, then:
$n = -1, 0, 3, 5$
$n = 1, 2, 4, 5$
$n = 0, 2, 4$
$n = -1, 1, 3, 5$
The sides of a triangle are $\sin \alpha ,\,\cos \alpha $ and $\sqrt {1 + \sin \alpha \cos \alpha } $ for some $0 < \alpha < \frac{\pi }{2}$. Then the greatest angle of the triangle is.....$^o$
If $\alpha ,\,\beta ,\,\gamma ,\,\delta $ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k$ , then the value of $4\sin \frac{\alpha }{2} + 3\sin \frac{\beta }{2} + 2\sin \frac{\gamma }{2} + \sin \frac{\delta }{2}$ is equal to
The equation $3{\sin ^2}x + 10\cos x - 6 = 0$ is satisfied, if
The number of solutions of $|\cos x|=\sin x$, such that $-4 \pi \leq x \leq 4 \pi$ is.
For $x \in(0, \pi)$, the equation $\sin x+2 \sin 2 x-\sin 3 x=3$ has