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

By tangent half angle identities we obtain by $t=\tan \frac x 2$


*

*$\cos 2x =2\cos^2 x-1=\frac{2(1-t^2)^2}{(1+t^2)^2}-1$

*$\cos 3x=4\cos^3 x-3\cos x=4\frac{(1-t^2)^3}{(1+t^2)^3}-3\frac{1-t^2}{1+t^2}$

*$\sin (4x)=2\sin (2x)\cos(2x)=4\sin x\cos x(2\cos^2 x-1)=4\frac{2t}{1+t^2}\frac{1-t^2}{1+t^2}\left(\frac{2(1-t^2)^2}{(1+t^2)^2}-1\right)$
then 
$$\cos 2x-\cos 3x+\sin 4x=0$$
$$\frac{2(1-t^2)^2}{(1+t^2)^2}-1-4\frac{(1-t^2)^3}{(1+t^2)^3}+3\frac{1-t^2}{1+t^2}+4\frac{2t}{1+t^2}\frac{1-t^2}{1+t^2}\left(\frac{2(1-t^2)^2}{(1+t^2)^2}-1\right)=0$$
$$2(1-t^2)^2(1+t^2)^2-(1+t^2)^4-4(1-t^2)^3(1+t^2)+$$$$+3(1-t^2)(1+t^2)^3+8t(1-t^2)(t^4-6t^2+1)=0$$
that is
$$t (t^7 - 4 t^6 - 9 t^5 + 28 t^4 - 5 t^3 - 28 t^2 + 5 t + 4) = 0$$
which confirms that $t=0$ is a solution then
$$t^7 - 4 t^6 - 9 t^5 + 28 t^4 - 5 t^3 - 28 t^2 + 5 t + 4=0$$
which seems to have others $5$ not trivial solutions.
