If both roots of quadratic equation ${x^2} + \left( {\sin \,\theta  + \cos \,\theta } \right)x + \frac{3}{8} = 0$ are positive and distinct then complete set of values of $\theta $ in $\left[ {0,2\pi } \right]$ is 

$2{\sin ^2}x + {\sin ^2}2x = 2,\, - \pi < x < \pi ,$ then $x = $

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

For each positive real number $\lambda$. Let $A_\lambda$ be the set of all natural numbers $n$ such that $|\sin (\sqrt{n+1})-\sin (\sqrt{n})|<\lambda$. Let $A_\lambda^c$ be the complement of $A_\lambda$ in the set of all natural numbers. Then,

The number of solutions of equation $3cos^2x - 8sinx = 0$ in $[0, 3\pi]$ is

Let $f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x$ $+C \cos 2 x+D \cos x+E$, and

$T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) \text {. }$Then, $T$