Solve: $\sqrt{\sqrt{3}\cot \theta }<1$.

The solution set of the equation $4\sin \theta ·\cos \theta -2\cos \theta -2\sqrt{3}\sin \theta +\sqrt{3}=0$ in the interval $(0, 2\pi )$ is

The general solution of the equation $\sin x+\sin 5x=\sin 2x+\sin 4x$ is

If $2{\cos }^2\left(\mathrm{\pi }+x\right)+3\sin \left(\pi +x\right)$ vanishes then the values of $x$ lying in the interval from $0$ to $2\mathrm{\pi }$ are

$\frac{\cos 3\theta }{2\cos 2\theta -1}=\frac{1}{2}$ if

If $\cos 2\theta +3\cos \theta =0,$ then

If $\sin \theta +7\cos \theta =5,$ then $\tan \left(\frac{\theta }{2}\right)$ is a root of the equation