Use the double-angle formula: \(\sin 2x=2\sin x\cos x\).

Then the equation becomes:

\(2\sin x\cos x=\sqrt{3}\sin x\).

Move everything to one side and factor out \(\sin x\):

\(\sin x(2\cos x-\sqrt{3})=0\).

Hence:

1) \(\sin x=0\Rightarrow x=\pi n\), \(n\in\mathbb Z\).

2) \(2\cos x-\sqrt{3}=0\Rightarrow \cos x=\dfrac{\sqrt{3}}{2}\Rightarrow x=2\pi n\pm\dfrac{\pi}{6}\), \(n\in\mathbb Z\).

Answer: \(x=\pi n\), or \(x=2\pi n\pm\dfrac{\pi}{6}\), \(n\in\mathbb Z\).

Let \(t=\tan x\). Then \(\cot x=\dfrac{1}{t}\).

We get the equation:

\(t-3=\dfrac{4}{t}\).

Multiply by \(t\):

\(t^2-3t-4=0\).

Factorize:

\((t-4)(t+1)=0\).

So \(t=4\) or \(t=-1\).

Hence:

\(\tan x=4\Rightarrow x=\arctan 4+\pi n\), \(n\in\mathbb Z\);

\(\tan x=-1\Rightarrow x=-\dfrac{\pi}{4}+\pi n\), \(n\in\mathbb Z\).

Write \(\sin 2x\) in terms of \(\sin x\) and \(\cos x\):

\(2\sin x\cos x+2\sin^2 x=0\).

Factor out \(2\sin x\):

\(2\sin x(\cos x+\sin x)=0\).

Hence:

1) \(\sin x=0\Rightarrow x=\pi n\), \(n\in\mathbb Z\).

2) \(\cos x+\sin x=0\Rightarrow \tan x=-1\Rightarrow x=-\dfrac{\pi}{4}+\pi n\), \(n\in\mathbb Z\).

Let \(t=\cos x\). Then we obtain the quadratic equation:

\(2t^2-t=1\Rightarrow 2t^2-t-1=0\).

Factorize:

\((2t+1)(t-1)=0\).

Therefore, \(t=1\) or \(t=-\dfrac12\).

Hence:

\(\cos x=1\Rightarrow x=2\pi n\), \(n\in\mathbb Z\);

\(\cos x=-\dfrac12\Rightarrow x=2\pi n\pm\dfrac{2\pi}{3}\), \(n\in\mathbb Z\).

Use the formula \(\sin 2x=2\sin x\cos x\):

\(2\sin x\cos x+\sqrt{2}\cos x=0\).

Factor out \(\cos x\):

\(\cos x(2\sin x+\sqrt{2})=0\).

Hence:

1) \(\cos x=0\Rightarrow x=\dfrac{\pi}{2}+\pi n\), \(n\in\mathbb Z\).

2) \(2\sin x+\sqrt{2}=0\Rightarrow \sin x=-\dfrac{\sqrt{2}}{2}\).

Then \(x=-\dfrac{\pi}{4}+2\pi n\) or \(x=\dfrac{5\pi}{4}+2\pi n\), \(n\in\mathbb Z\).