Solve $\cos 2x-\cos 3x+\sin 4x = 0$

Question

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

Answer 1

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.

Answer 2

Factorirzing your given equation we get $$2 \sin \left(\frac{x}{2}\right) \left(\sqrt{2} \sin \left(\frac{5 x}{2}+\frac{\pi }{4}\right)+\cos \left(\frac{x}{2}\right)+\cos \left(\frac{3 x}{2}\right)+\cos \left(\frac{7 x}{2}\right)\right)=0$$

Answer 3

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 $$\left( \begin{array}{cc} 0 & 1.00000 \\ 1 & 0.95437 \\ 2 & 0.95541 \end{array} \right)$$

$$\left( \begin{array}{cc} 0 & 2.70000 \\ 1 & 2.72792 \\ 2 & 2.72687 \\ 3 & 2.72687 \end{array} \right)$$

$$\left( \begin{array}{cc} 0 & 3.80000 \\ 1 & 3.84203 \\ 2 & 3.84161 \end{array} \right)$$

$$\left( \begin{array}{cc} 0 & 4.90000 \\ 1 & 4.85398 \\ 2 & 4.85570 \\ 3 & 4.85571 \end{array} \right)$$

$$\left( \begin{array}{cc} 0 & 5.60000 \\ 1 & 5.66083 \\ 2 & 5.66212 \\ 3 & 5.66213 \end{array} \right)$$