<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] Using the tangent half-angle substitution, you end with
$$\left(1+\sqrt{2}\right) t^4+2 \left(1+\sqrt{2}\right) t^2-8 t+(1+\sqrt{2})=0$$By inspection (with all these $1$'s and $\sqrt{2}$'s),  $\color{red}{t=(\sqrt{2}-1)}$ is a root.
Factoring, we are left with
$$\left(1+\sqrt{2}\right) t^3+t^2+\left(1+3 \sqrt{2}\right) t-(3+2 \sqrt{2})=0$$ for which $$\Delta=-64 \left(63+44 \sqrt{2}\right) \quad <0$$ Using the hyperbolic method for only one real root which is (!!)
$$\color{red}{t=\frac{1}{3} \left(1-\sqrt{2}+4 \sqrt{3-\sqrt{2}} \sinh \left(\frac{1}{3}
   \sinh ^{-1}\left(\frac{5}{14} \sqrt{\frac{1}{7} \left(173+81
   \sqrt{2}\right)}\right)\right)\right)}$$ I shall not try to find what is this number which is not recognized by inverse symbolic calculators.
To the first root corresponds
$$x_1=2 \tan ^{-1}\left(\sqrt{2}-1\right)=\frac \pi 4+2k\pi$$
and the second is
$$x_2\sim 2\tan ^{-1}(0.778669) \sim 1.323197$$
</s>