<s>[INST] Solve $\cos 2x-\cos 3x+\sin 4x = 0$ Solve Equation$$\cos 2x-\cos 3x+\sin 4x=0 $$
Attemp:Developping up to $\sin x,\cos x$, equation is :
$(\cos x-1)(4\cos^2x+2\cos x-1)=4\sin x(2\cos^3x-\cos x)$
Note that squaring will give solutions $x_i$ such that either $x_i$, either $-x_i$ is solution of the current equation.
So $(\cos x-1)^2(4\cos^2x+2\cos x-1)^2=16(1-\cos^2x)(2\cos^3x-\cos x)^2$
We obviously get $\cos x=1$ (which indeed is a solution of original equation) and it remains
$(1-\cos x)(4\cos^2x+2\cos x-1)^2=16(1+\cos x)(2\cos^3x-\cos x)^2$
Setting $\cos x=y$, this is :
$(1-y)(4y^2+2y-1)^2=16(1+y)(2y^3-y)^2$
Which is $64y^7+64y^6-48y^5-64y^4-4y^3+16y^2+5y-1=0$
I found no clever way to solve this degree-7 polynomial
 [/INST] Factorirzing your given equation we get
$$2 \sin \left(\frac{x}{2}\right) \left(\sqrt{2} \sin
   \left(\frac{5 x}{2}+\frac{\pi }{4}\right)+\cos
   \left(\frac{x}{2}\right)+\cos \left(\frac{3
   x}{2}\right)+\cos \left(\frac{7 x}{2}\right)\right)=0$$
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