<s>[INST] Solve for $x$ in $ \sin x + \sin x \cos x - (\frac{1}{\sqrt{2}} + \frac{1}{2})= 0 $ How can to solve for $x$ in $$ \sin x + \sin x \cos x - \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right)= 0.$$
My try:
$$ \sin x + \sin x \cos x = \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right) $$
$$  (\sin x + \sin x \cos x)^2 = (\frac{1}{\sqrt{2}} + \frac{1}{2})^2
$$
$$ \sin^2x + 2\sin^2 x \cos x + \sin^2x \cos^2x - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 =  0
$$
$$ \sin^2 x( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ (1 - \cos^2x)( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ 1 + 2 \cos x + \cos^2 x - \cos^2 x - 2 \cos^3 x - \cos^4 x- (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
thx for all ur help
 [/INST] You may continue with
$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
knowing that $ \cos x = \frac1{\sqrt2} $ is a solution from inspection, as commented. So, factorize  the equation as
$$ (\cos x - \frac1{\sqrt2})(\cos^3x +(2+\frac1{\sqrt2})\cos^2x+ (\frac12+\sqrt2)\cos x-1+\frac1{2\sqrt2})=0
$$
where the cubic polynomial has one real root that can be calculated analytically with the Cardano formula.
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