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

I think that this has to be solved numerically using Newton method. Graphing for $0 \leq x \leq 2\pi$, we can notice, beside the trivial roots $x=0$ and $x=2\pi$, solutions close to $1.0$, $2.7$, $3.8$, $4.9$ and $5.6$.
Using these as starting points, Newton iterates are 
$$\left(
\begin{array}{cc}
 0 & 1.00000 \\
 1 & 0.95437 \\
 2 & 0.95541
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 0 & 2.70000 \\
 1 & 2.72792 \\
 2 & 2.72687 \\
 3 & 2.72687
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 0 & 3.80000 \\
 1 & 3.84203 \\
 2 & 3.84161
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 0 & 4.90000 \\
 1 & 4.85398 \\
 2 & 4.85570 \\
 3 & 4.85571
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 0 & 5.60000 \\
 1 & 5.66083 \\
 2 & 5.66212 \\
 3 & 5.66213
\end{array}
\right)$$
