<s>[INST] How to solve $y' + y^2 - 2y\sin x + \sin^2x = \cos x$ How to solve the following equation?
$$y' + y^2 - 2y\sin x + \sin^2x = \cos x$$
It is necessary to determine the type and total solution.
Help me please.
 [/INST] $$y' + y^2 - 2y\sin x + \sin^2x = \cos x$$
$$(y' -  \cos x) + (y-\sin x)^2 = 0$$
Say $z=y- \sin x$ 
Then $\frac{dz}{dx}= y'-\cos x$
So we have $$\frac{dz}{dx}+z^2=0$$
$$\int z^{-2} dz = -\int dx$$
$$\frac{1}{z}=x+c$$
$$\frac{1}{y- \sin x}=x+c$$
$$y = \sin x + \frac{1}{x+c}$$
This is the required solution.
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