<s>[INST] Finding the value of $\sin^3x+\cos x$, if $x$ is an acute angle satisfying $2\sin x\sin\left(\frac{x}{2}\right)=1-\sin x$ Given that $x$ is an acute angle satisfying $$2\sin x\sin\left(\frac{x}{2}\right)=1-\sin x$$
Then find the exact value of $$\sin^3 x+\cos x$$
My try:
Letting $x=2t$ we get
$$2\sin(2t)\sin(t)=1-\sin(2t)=(\cos t-\sin t)^2$$
So we get
$$2\sin t\sqrt{\cos t}=\cos t-\sin t$$
But i am stuck here
 [/INST] If there is no typo, there is no analytical solution to the first equation.
For example, if you let $x=4\tan^{-1}(t)$, you would need to solve for $t$ the sextic polynomial
$$t^6+4 t^5+19 t^4-13 t^2-4 t+1=0$$ which does not factor.
Beside purely numerical methods, we can try series expansions. Graphing or by inspection, for an acute angle, the solution is just above $\frac \pi 5$. Using the first term of the series expansion, an approximation is
$$x=\frac \pi 5+\frac{4-\sqrt{2 \left(5+\sqrt{5}\right)}}{3+2 \sqrt{5}}=0.654519$$ while the solution given by Newton method is $0.654434$ as @Suzu Hirose already reported.
Edit
Using my favored, $1,400$ years old, approximation of the sine function
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ we end with a quartic equation (better than the sextic), namely,
$$108 x^4-324 \pi  x^3+111 \pi ^2 x^2+110 \pi ^3 x-25 \pi ^4=0$$
Letting $x=\pi\,y$
$$108 y^4-324 y^3+111 y^2+110 y-25=0$$
which shows four real solutions which express in terms of radicals; the one we look for is the smallest positive root.
Just for the fun, the solution is
$$36\,y=27 -\sqrt{3} \sqrt{169+a+\frac{9649}{a}}+\sqrt{3}
   \sqrt{338-a-\frac{1056
   \sqrt{3}}{\sqrt{169+a+\frac{9649}{a}}}-\frac{9649}{a}}$$ where
$$a=\sqrt[3]{450793+480 i \sqrt{3017094}}$$ which gives
$$y=0.208264 \quad \implies \quad x=0.654282$$ which is quite good for an approximation (relative error of $0.023$%).
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