<s>[INST] $ \sin{x}\ +\ \frac{1}{2}\sin{2x}\ +\ \frac{1}{3}\sin{3x}\ +\ \frac{1}{4}\sin{4x}\ =\ \frac{2}{3}\left(\cos{x}+1\right)\left(\sin^5x\ +\ 4\right)$ SORRY IF MY TITTLE IS UNCLEAR WITH ONLY MATH FUNCTIONS. IT'S MORE THAN 150 CHARACTERS


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*This is my math problem


$$ \sin {x}\ +\ \frac{1}{2}\sin {2x}\ +\ \frac{1}{3}\sin {3x}\ +\ \frac{1}{4}\sin {4x}\ =\ \frac{2}{3}\left(\cos {x}+1\right)\left(\sin ^5x\ +\ 4\right) \left(*\right)
$$


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*This is my effort


$$ \left(*\right) <=> 12\sin {x\left(1+\cos x\right)\ +\ 12\sin {x}\ -\ 16\sin ^3x\ +\ 12\sin {\left(x\right)\cos {\left(x\right)}\cos {\left(2x\right)}}}\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)
$$
$$   <=> 12\sin {x\left(1+\cos x\right)\ + 4\sin {x\left(2\cos {\left(2x\right)\left(1+\cos {x}\right)\\ +\ \left(1+\cos {x}\right)\left(2\cos ^2x\ -\ 2\cos {x\ +\ 1}\right)}\right)}}\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right) 
$$
$$ <=> 4\sin {x\left(1+\cos {x}\right)\left(6\cos ^2x\ -\ 2\cos {x\ +\ 2}\right)\ =\ 8\left(\cos x\ +\ 1\right)\left(\sin ^5x\ +\ 4\right)}
$$
Now 
$$\cos {x} = -1
$$
or
$$ 6\sin {x}\cos ^2x\ -\ 2\cos {x}\sin {x}\ +\ 2\sin {x}\ -\ 2\sin ^5x\ -\ 8\ =\ 0 
$$
To here I tried many ways like converting all to \sin , group somes together but it didn't work. Please give me some HINTS
 [/INST] An algorithmic approach for many of these problems can be:
You can call $t=\tan(\frac{x}{2})$. Then $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t}{1+t^2}$. Making this substitution your equation 
$$6\sin(x)cos^2(x) − 2\cos(x)\sin(x) + 2\sin(x) − 2\sin^5(x) − 8 = 0$$
becomes a polynomial equation.
$$6(2t)(1-t)^2(1+t^2)^2− 2(1-t)(2t)(1+t^2)^3 + 2(2t)(1+t^2)^4−2(2t)^5 − 8(1+t^2)^5=0$$
With Sturm's theorem we check that this polynomial doesn't have real solutions.
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