How to solve the trigonometric equation $\cos17x=20\cos x$? How to solve the following trigonometric equation?
$$\cos17x=20\cos x$$
I'm really awful in trigonometry. I tried division of both sides by $20$. Thanks.
 A: The function $f(x):=20\cos x-\cos(17x)$ is $2\pi$-periodic and even; therefore it suffices to analyze what happens in the $x$-interval $[0,\pi]$. Write $x:={\pi\over2}-y$ and consider  the function
$$g(y):=f\left({\pi\over2}-y\right)=20\sin y-\sin(17 y)\qquad\left(-{\pi\over2}\leq y\leq{\pi\over2}\right)$$
instead. Since $g$ is odd we can even restrict to $0\leq y\leq{\pi\over2}$.
When $0<y\leq{1\over10}$  truncating the $\sin$ series after a negative term provides  the estimate
$$\sin y>y-{y^3\over6}= y\left(1-{y^2\over6}\right)>{9\over10}y\ .$$
It follows that
$$20\sin y-\sin(17y)> 18 y-17 y>0\qquad\left(0<y\leq{1\over10}\right)\ .$$
Furthermore one has
$$20\sin y-\sin(17y)\geq 20\sin{1\over10}-1>20{9\over10}{1\over10}-1={4\over5}>0\qquad\left({1\over10}\leq y\leq{\pi\over2}\right)\ .$$
It follows that the only zero of $g$ is at $y=0$, which corresponds to $x={\pi\over2}$. Therefore the solutions of the original equation are the odd multiples of ${\pi\over2}$.
A: Hint: The key to a simple solution seems to me to be 
First to identify where $-1\le 20\cos x \le 1$ because this is necessary for a solution.
Then show that the curve of $20\cos x$ is steeper than that of $\cos 17 x$ in those intervals (e.g. use the derivative.)
Then show that the solutions where both expressions are simultaneously zero are the only ones (and pin these down)
Here a very rough sketch shows what is going on, and avoids a lot of unnecessary algebra.
A: There probably is a simpler solution, but ... you can use Chebyshev polynomials : if $c=\cos(x)$, your equation becomes
$T_{17}(c)=20c$, or
$$
c(65536c^{16} - 278528c^{14} + 487424c^{12} - 452608c^{10} + 239360c^8 - 71808c^6 + 11424c^4 - 816c^2 - 3)=0
$$
But the polynomial $A=65536c^{16} - 278528c^{14} + 487424c^{12} - 452608c^{10} + 239360c^8 - 71808c^6 + 11424c^4 - 816c^2 - 3$ has only two opposite roots, both of absolute value
$>1$. So $c=0$, and $x=\pm \frac{\pi}{2}+2n\pi$ as Macavity guessed.
A: This not a good approach, but
$$\cos 17x = \\17\cos x \sin^{16}x-680\cos^3x sin^{14}x+6188\cos^5x \sin^{12}x-19448\cos^7X\sin^{10}x+24310\cos^9x\sin^8x-12376\cos^{11}x\sin^6x+2380\cos^13x\sin^4x-136\cos^{15}x\sin^2x+\cos^{17}x$$
I have pick this from here.
