The equation $2{\cos ^2}\left( {\frac{x}{2}} \right)\,{\sin ^2}x\, = \,{x^2}\, + \,\frac{1}{{{x^2}}},\,0\,\, \leqslant \,\,x\,\, \leqslant \,\,\frac{\pi }{2}\,\,$ has

The only value of $x$ for which ${2^{\sin x}} + {2^{\cos x}} > {2^{1 - (1/\sqrt 2 )}}$ holds, is

The sum of all $x \in[0, \pi]$ which satisfy the equation $\sin x+\frac{1}{2} \cos x=\sin ^2\left(x+\frac{\pi}{4}\right)$ is

Let $S=\left\{\theta \in(0,2 \pi): 7 \cos ^{2} \theta-3 \sin ^{2} \theta-2\right.$ $\left.\cos ^{2} 2 \theta=2\right\}$. Then, the sum of roots of all the equations $x ^{2}-2\left(\tan ^{2} \theta+\cot ^{2} \theta\right) x +6 \sin ^{2} \theta=0$ $\theta \in S$, is$...$

If $5{\cos ^2}\theta + 7{\sin ^2}\theta - 6 = 0$, then the general value of $\theta $ is

All the pairs $(x, y)$ that satisfy the inequality ${2^{\sqrt {{{\sin }^2}{\kern 1pt} x - 2\sin {\kern 1pt} x + 5} }}.\frac{1}{{{4^{{{\sin }^2}\,y}}}} \leq 1$ also Satisfy the equation