Solution of Trigonometric Equations

Total number of solutions of $\sin x.\tan 4x=\cos x$ belonging to $\left(0,\pi \right)$ are 

The value '$a$' for which the equation $4{cosec}^2\left(\pi \left(a+x\right)\right)+a^2-4a=0$ has a real solution is:

$\cos 2x-3\cos x+1=$ $\frac{1}{\left(\cot 2x-\cot x\right)\sin \left(x-\pi \right)}$ holds if : 

If $\sin x=\cos y, \sqrt{6}\sin y=\tan z$ and $2\sin z=\sqrt{3}\cos x; u, v, w$ denotes respectively ${\sin }^{2}x,{\sin }^{2}y,{\sin }^{2}z$, then the value of the triplet $\left(u,v,w\right)$ is

$\tan \left(\frac{p\mathrm{\pi }}{4}\right)=\cot \left(\frac{q\pi }{4}\right)$ if

$\mathrm{}$$6{\tan }^{2}x-2{\cos }^{2}x=\cos 2x$ (where $x\ne \frac{\mathrm{\pi }}{2}$) if:

The number of solutions of the equation, ${\sin }^{5}x-{\cos }^{5}x=\frac{1}{\cos x}-\frac{1}{\sin x}\left(\sin x\ne \cos x\right)$ is :

The general solution of the equation, $\frac{1-\sin x+\cdots +(-1{)}^n{\sin }^{n}x+\cdots }{1+\sin x+\cdots +{\sin }^{n}x+\cdots }=\frac{1-\cos 2x}{1+\cos 2x}$ is 

The number of all possible $5$ -triplets $\left(a_1,a_2,a_3,a_4,a_5\right)$ such that $a_1+a_2\sin x$ $+a_3\cos x+a_4\sin 2x+a_5\cos 2x=0$ holds for all $x$ is :

The number of all possible triplets $\left(a_1,a_2,a_3\right)$ such that $a_1+a_2\cos 2x+a_3{\sin }^{2}x=0$ for all $x$ is :

The equation ${\sin }^{4}x+{\cos }^{4}x+\sin 2x+\alpha =0$ is solvable for:

The general solution of the equation; $\cos \left(x\right)\cos \left(6x\right)=-1$  is

The most general value of $\theta$ which satisfy both the equations $\cos \theta =-\frac{1}{\sqrt{2}}$ and $\tan \theta =1,$ is :

Consider the equation $\sin 3\alpha =4\sin \alpha .\sin (x+\alpha )\sin (x-\alpha ),\alpha \in (0,\pi )$ and $x\in (0,2\pi ),$ then for $x$

Total number of solutions of $\left[\sin x\right]+\cos x=0$; where $[.]$ denotes the greatest integer function, for $x\in [0,2\pi ]$, is :

If the equation $1+{\sin }^{2}x\theta =\cos \theta$ has a non-zero solution in $\theta ,$ then$x$ must be:

Total number of solutions of the equation $2x=3\pi (1-\cos x),$ is :

The number of solutions of the equation;  $x^3+2x^2+5x+2\cos x=0$ in $[0,2\pi ]$ is:

Solve, $\sin x+\sin \left(\frac{\pi }{8}\sqrt{(1-\cos 2x{)}^2+{\sin }^{2}2x}\right)=0, x\in \left[\frac{5\pi }{2},\frac{7\pi }{2}\right]$

Solve the equation:

$\tan \left(\frac{\pi }{2}\cos \theta \right)=\cot \left(\frac{\pi }{2}\sin \theta \right)$