$ \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


*

*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
 A: 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.
A: It remains to solve the following equation.
$$\sin{x}(3\cos^2x-\cos{x}+1)=\sin^5x+4.$$
We'll prove that
$$\sin{x}(3\cos^2x-\cos{x}+1)<\sin^5x+4,$$ which says that this equation has no real roots.
Indeed,
$$\sin{x}(3\cos^2x-\cos{x}+1)=\sin{x}(3(1-\sin^2x)+1)-\sin{x}\cos{x}\leq$$
$$\leq\sin{x}(4-3\sin^2x)+\frac{\sin^2x+\cos^2x}{2}=4\sin{x}-3\sin^3x+\frac{1}{2}.$$
Now, let $\sin{x}=a$ and since $3\cos^2x-\cos{x}+1>0$ and $\sin^5x+4>0,$ we see that $a>0$
and we need to prove that
$$a^5+3a^3-4a+3.5>0,$$ which is true by AM-GM:
$$a^5+3a^3-4a+3.5>3a^3+2\cdot1.75-4a\geq\left(3\sqrt[3]{3\cdot1.75^2}-4\right)a>0.$$
I used the following AM-GM:

for non-negatives $x$, $y$ and $z$ we have:
  $$\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$$ or
  $$x+y+z\geq3\sqrt[3]{xyz}.$$

Here $x=3a^3$ and $y=z=1.75.$
