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

Use the formula \(\cos 2a=\cos^2 a-\sin^2 a\):

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

That is,

\(2\cos 4x=\sqrt{3}\Rightarrow \cos 4x=\dfrac{\sqrt{3}}{2}\).

Then

\(4x=2\pi n\pm\dfrac{\pi}{6}\), \(n\in\mathbb Z\).

Divide by 4:

\(x=\dfrac{\pi n}{2}\pm\dfrac{\pi}{24}\), \(n\in\mathbb Z\).

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

We get:

\(t+1=\dfrac{2}{t}\).

Multiply by \(t\):

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

Factorize:

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

So \(t=1\) or \(t=-2\).

Hence:

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

\(\tan x=-2\Rightarrow x=-\arctan 2+\pi n\), \(n\in\mathbb Z\).

Factor out \(\sin x\):

\(\sin x(\cos x-\sin x)=0\).

Hence:

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

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

Move \(\cos 5x\) to the right:

\(\sin 5x=-\cos 5x\).

Divide by \(\cos 5x\) (when \(\cos 5x=0\), the left side is not zero):

\(\tan 5x=-1\).

Then

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

Therefore,

\(x=-\dfrac{\pi}{20}+\dfrac{\pi n}{5}\), \(n\in\mathbb Z\).

From the equation

\(\tan^2 2x=3\)

we get

\(\tan 2x=\pm\sqrt{3}\).

Then

\(2x=\pm\dfrac{\pi}{3}+\pi n\), \(n\in\mathbb Z\).

Divide by 2:

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

Let \(t=\sin x\). Then

\(2t^2-5t+2=0\).

Factorize:

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

So \(t=\dfrac12\) or \(t=2\).

But \(\sin x\) cannot be equal to 2, so only \(\sin x=\dfrac12\) remains.

Hence

\(x=\dfrac{\pi}{6}+2\pi n\) or \(x=\dfrac{5\pi}{6}+2\pi n\), \(n\in\mathbb Z\).

Since \(\dfrac{\sin x}{\cos x}=\tan x\), we get:

\(\tan^2 x+4\tan x=5\).

Let \(t=\tan x\). Then:

\(t^2+4t-5=0\).

Factorize:

\((t-1)(t+5)=0\).

Therefore, \(t=1\) or \(t=-5\).

Hence:

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

\(\tan x=-5\Rightarrow x=-\arctan 5+\pi n\), \(n\in\mathbb Z\).

Move 1 to the right-hand side:

\(2\sin x\ge 1\Rightarrow \sin x\ge \dfrac12\).

On the unit circle, this means that the angle \(x\) lies between \(\dfrac{\pi}{6}\) and \(\dfrac{5\pi}{6}\), including the endpoints.

Taking the period \(2\pi\) into account, we get:

\(x\in\left[\dfrac{\pi}{6}+2\pi n,\;\dfrac{5\pi}{6}+2\pi n\right],\; n\in\mathbb Z\).

Use the double-angle formula:

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

Then the inequality becomes:

\(\sin 2x<0\).

Sine is negative on the intervals \((\pi+2\pi n,\;2\pi+2\pi n)\).

Therefore,

\(2x\in(\pi+2\pi n,\;2\pi+2\pi n)\).

Divide by 2:

\(x\in\left(\dfrac{\pi}{2}+\pi n,\;\pi+\pi n\right),\; n\in\mathbb Z\).

Transform the inequality:

\(\sqrt{3}\tan x+3>0\Rightarrow \sqrt{3}\tan x>-3\Rightarrow \tan x>-\sqrt{3}\).

On each interval \(\left(-\dfrac{\pi}{2}+\pi n,\;\dfrac{\pi}{2}+\pi n\right)\), the function \(\tan x\) is increasing.

The tangent takes the value \(-\sqrt{3}\) at \(x=-\dfrac{\pi}{3}+\pi n\).

Therefore, the solution is:

\(x\in\left(-\dfrac{\pi}{3}+\pi n,\;\dfrac{\pi}{2}+\pi n\right),\; n\in\mathbb Z\).

Use the formula:

\(\cos^2 x-\sin^2 x=\cos 2x\).

Then the inequality becomes:

\(\cos 2x\ge 0\).

Cosine is non-negative on the intervals

\(2x\in\left[-\dfrac{\pi}{2}+2\pi n,\;\dfrac{\pi}{2}+2\pi n\right],\; n\in\mathbb Z\).

Divide by 2:

\(x\in\left[-\dfrac{\pi}{4}+\pi n,\;\dfrac{\pi}{4}+\pi n\right],\; n\in\mathbb Z\).