(sin2x)^cosx equation

Given the equation
$$\sin^{\cos{\left(x \right)}}{\left(2 x \right)} = 0$$
transform
$$\sin^{\cos{\left(x \right)}}{\left(2 x \right)} - 1 = 0$$
$$\sin^{\cos{\left(x \right)}}{\left(2 x \right)} - 1 = 0$$
Do replacement
$$w = \cos{\left(x \right)}$$
$$\sin^{w}{\left(2 x \right)} - 1 = 0$$
or
$$\sin^{w}{\left(2 x \right)} - 1 = 0$$
or
$$\sin^{w}{\left(2 x \right)} = 1$$
or
$$\sin^{w}{\left(2 x \right)} = 1$$
- this is the simplest exponential equation
Do replacement
$$v = \sin^{w}{\left(2 x \right)}$$
we get
$$v - 1 = 0$$
or
$$v - 1 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 1$$
We get the answer: v = 1
do backward replacement
$$\sin^{w}{\left(2 x \right)} = v$$
or
$$w = \frac{\log{\left(v \right)}}{\log{\left(\sin{\left(2 x \right)} \right)}}$$
The final answer
$$w_{1} = \frac{\log{\left(1 \right)}}{\log{\left(\sin{\left(2 x \right)} \right)}} = 0$$
do backward replacement
$$\cos{\left(x \right)} = w$$
Given the equation
$$\cos{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
$$x_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$x_{1} = \pi n + \operatorname{acos}{\left(0 \right)}$$
$$x_{1} = \pi n + \frac{\pi}{2}$$
$$x_{2} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$x_{2} = \pi n - \pi + \operatorname{acos}{\left(0 \right)}$$
$$x_{2} = \pi n - \frac{\pi}{2}$$