Solve : $$\cos{x} \cos{12x} \cos{15x} = \cos{5x} \cos{7x} \cos{16x}$$
My turn: L.H.S $$= \frac{1}{8}\frac{\sin{8x \cos{12x}\cos{15x}}}{\sin{x} \cos{2x} \cos{4x}} $$ I tried this hopefully that i can get some identical terms to simplify the questions but it did not make sense
On
Given
$$\cos{x} \cos{12x} \cos{15x} = \cos{5x} \cos{7x} \cos{16x}$$
we can expand the LHS and RHS through
$$\cos{x}\cos{y}=\frac{1}{2}\big[\cos(x−y)+\cos(x+y)\big]$$
to form
$$\frac{1}{4} \big(\cos{2 x} + \cos{4 x} + \cos{26 x} + \cos{28 x})=\frac{1}{4} \big(\cos{4x} + \cos{14 x} + \cos{18 x} + \cos{28 x}\big)$$
which reduces to
$$\cos{2 x} + \cos{26 x}=\cos{14 x} + \cos{18 x}\tag{1}$$
then from
$$\cos{x}+\cos{y}=2\cos\Big(\frac{x+y}{2}\Big)\cos\Big(\frac{x-y}{2}\Big)$$
and
$$\cos(x)=\cos(-x)$$
we have
$$2\cos{14x}\cos(-12x)= 2\cos{16x}\cos(-2x)$$ $$\cos{14x}\cos{12x}= \cos{16x}\cos{2x}\tag{2}$$
At this point, I don't see a way to reduce $(1)$ or $(2)$.
On
If you don't mind a little algebra and don't want to be clever, you can simply let $z = e^{i x}$ and then note that $2 \cos(n x) = z^n + z^{-n}$, reducing the problem to a polynomial equation in $z$. And indeed
$$(z + z^{-1})(z^{12} + z^{-12})(z^{15} + z^{-15}) - (z^5 + z^{-5})(z^7 + z^{-7})(z^{16} + z^{-16})$$ $$ = z^{-26} (z^{52} - z^{44} - z^{40} + z^{28} + z^{24} - z^{12} - z^8 + 1)$$ $$ = z^{-26} (z-1)^2 (z+1)^2 (z^2 + 1)^2 (z^4 + 1)(z^4 - z^3 + z^2 - z + 1)(z^4 + z^3 + z^2 + z + 1)$$ $$\times \ (z^8 - z^6 + z^4 - z^2 + 1)(z^{24} - z^{12} + 1)$$ $$= \frac{(z^{18} + z^{-18})(z^{10} - z^{-10})(z^4 - z^{-4})}{z^6 + z^{-6}} = -4 \cdot \frac{\cos(18x)}{\cos(6 x)} \cdot \sin(10 x) \cdot \sin(4 x)$$
The first two lines are just algebra. The next line uses the "accident" that the factors are all cyclotomic, so the answer can be written in terms of cyclotomic polynomials, which is done on the final line (this won't happen for general choices of parameters). The solutions are of course periodic modulo $\pi$. The polynomial has a separable (no repeated factors) part of degree $44$, and then the roots $-1,1,i,-i$ each with multiplicity two. In particular, there are $48$ distinct roots and consequently $48$ distinct solutions for $x \in [0,2 \pi)$ and $24$ distinct solutions for $x$ in $[0,\pi)$. They turn out to have nice expressions in terms of degrees given as follows:
$$0,5, 18, 25, 35, 36, 45, 54, 55, 65, 72, 85, 90,$$ $$ 95, 108, 115, 125, 126, 135, 144, 145, 155, 162, 175.$$
Note that the answer is nice (rational degrees) exactly because the difference in factors was purely cyclotomic.
This example actually has interesting connections to $S$-unit equations in cyclotomic fields but that is a little too far afield to really discuss here.
On
Not very nice.
Starting from Axion004's equation $(1)$,we have to solve $$\cos (2 x)-\cos (14 x)-\cos (18 x)+\cos (26 x)=0$$ Let $t=\cos (2 x)$ and expand to get $$4096 t^{13}-13312 t^{11}+16384 t^9-9472 t^7+2592 t^5-300 t^3+12 t=0$$ which can factorize as $$t(t^2-1)(4096 t^{10}-9216 t^8+7168 t^6-2304 t^4+288 t^2-12)=0$$ For the last factor, let $t^2=z$ to get $$1024 z^5-2304 z^4+1792 z^3-576 z^2+72 z-3=0$$ Two of the roots (not easy to find them) are $$z=\frac{3\pm\sqrt{5}}{8} $$ So, we are left with $$64 z^3-96 z^2+36 z-3=0$$ which can be solved using the trigonometric method; this gives as roots $$z=\frac{1}{2} \left(1+\cos \left(\frac{\pi }{9}-\frac{2 \pi k}{3}\right)\right) \qquad \text{with} \qquad k=0,1,2$$
I let you the fun to go back to $t$ and then to $x$.
Possible direction:
\begin{align} 2\cos x \cos 15x & \equiv \cos 14x +\cos 16x \\ 2\cos 5x \cos 7x & \equiv \cos 2x +\cos 12x \\ (\cos 14x+\cos 16x) \cos 12x &= (\cos 2x+\cos 12x) \cos 16x \\ \cos 12x \cos 14x &= \cos 2x \cos 16x \\ \end{align}