If $2 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=0$ has exactly 3 solutions in the interval $\left[0, \frac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}$, then the roots of the equation $x^2+\mathrm{n} x+(\mathrm{n}-3)=0$ belong to :

If $\sqrt 3 ({\cos ^2}x) = (\sqrt 3 - 1)\cos x + 1$, the number of solutions of the given equation when $x \in \left[ {0,{\pi \over 2}}…

The number of solutions of the equation $|\cot x| = \cot x + {1 \over {\sin x}}$ in the interval [ 0, 2$\pi$ ] is

Let $$S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\}$$ and $\beta=\sum_\limits{x…

Let $$\mathrm{S = \{ \theta \in [0,2\pi ):\tan (\pi \cos \theta ) + \tan (\pi \sin \theta ) = 0\}}$$. Then $$\sum\limits_{\theta \in S}…

If m and n respectively are the numbers of positive and negative values of $\theta$ in the interval $[-\pi,\pi]$ that satisfy the equation…