$\displaystyle cos5x+cosx-sin4x=0$
Do you have any idea to simplify and find a general solution for it.
First of all, $\displaystyle \displaystyle \begin{align*} \cos{5x} \end{align*}$ is not $\displaystyle \displaystyle \begin{align*} 2\cos{3x}\cos{2x} \end{align*}$. If you split it up as $\displaystyle \displaystyle \begin{align*} \cos{(3x + 2x)} \end{align*}$ you will need to use this identity: $\displaystyle \displaystyle \begin{align*} \cos{(\alpha + \beta )} \equiv \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta} \end{align*}$. This is the angle sum identity I mentioned in my first post.
It also appears that the $\displaystyle \displaystyle \begin{align*} \cos{x} \end{align*}$ term has disappeared in the second line.
Let z=x+i y be a complex number. Substitute cos(5x), cos(x), sin(4x) with z component and solve for z...then arguments of z solutions are solutions to f(x) as follows:
$\displaystyle f(z)=\frac{1}{2}\left(z^5+\frac{1}{z^5} \right)+ \frac{1}{2} \left(z+\frac{1}{z}\right)-\frac{1}{2i}\left(z^4-\frac{1}{z^4}\right)$
f(z) simplifies to:
$\displaystyle f(z)=\frac{(-i+z) (i+z) \left(-i+z^2\right) \left(i+z^2\right) \left(1-i z-z^2+i z^3+z^4\right)}{2 z^5} =0$ z is not 0
$\displaystyle z= \left\{-i,i,-(-1)^{1/4},(-1)^{1/4},-(-1)^{3/4},(-1)^{3/4},-\frac{1}{4} i \left(1+\sqrt{5}-i \sqrt{2 \left(5-\sqrt{5}\right)}\right),-\frac{1}{4} i \left(1+\sqrt{5}+i \sqrt{2 \left(5-\sqrt{5}\right)}\right),\frac{1}{4} i \left(-1+\sqrt{5}-i \sqrt{2 \left(5+\sqrt{5}\right)}\right),\frac{1}{4} i \left(-1+\sqrt{5}+i \sqrt{2 \left(5+\sqrt{5}\right)}\right)\right\}$
Arguments of z are as follows:
$\displaystyle x=\left\{-\frac{\pi }{2},\frac{\pi }{2},-\frac{3 \pi }{4},\frac{\pi }{4},-\frac{\pi }{4},\frac{3 \pi }{4},-\pi -\text{ArcTan}\left[\frac{-1-\sqrt{5}}{\sqrt{2 \left(5-\sqrt{5}\right)}}\right],\text{ArcTan}\left[\frac{-1-\sqrt{5}}{\sqrt{2 \left(5-\sqrt{5}\right)}}\right],\text{ArcTan}\left[\frac{-1+\sqrt{5}}{\sqrt{2 \left(5+\sqrt{5}\right)}}\right],\pi -\text{ArcTan}\left[\frac{-1+\sqrt{5}}{\sqrt{2 \left(5+\sqrt{5}\right)}}\right]\right\}$
$\displaystyle \cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
so
$\displaystyle \cos 5x+\cos x=2\cos 3x\cos 2x.$
That allows the equation to be written as
$\displaystyle 2\cos 2x(\cos 3x - \sin 2x)=0.$
Now use the identities
$\displaystyle \cos 3x = 4\cos^{3}x-3\cos x$ and $\displaystyle \sin 2x = 2\sin x \cos x,$
remove $\displaystyle \cos x$ as a factor and turn what's left inside the brackets into a quadratic in $\displaystyle \sin x.$