So I am asked to find the sum of all solutions to
$\sin(3x) = \cos(6x)$, $x\in[0,2\pi)$
I'm trying to turn everything in terms of sine, so I do:
$\sin(3x) = 1-2\sin^2(6x)$
However, my book says the correct way to begin this is:
$\sin(3x) = 1-2\sin^2(3x)$
Why is it $1-2\sin^2(3x)$ instead of $1-2\sin^2(6x)$ if I want it to be equivalent to $\cos(6x)$? I thought the argument stays the same when you convert? Very confused.
On
Why is it $1−2\sin^2(3x)$ instead of $1−2\sin^2(6x)$ if I want it to be equivalent to $\cos(6x)$?
One can rewrite $\cos(6x)$ as $\cos(3x+3x)$. By the trigonometric addition formula for cosine,
$$\cos(3x+3x)=\cos(3x)\cos(3x)-\sin(3x)\sin(3x)=\cos^2(3x)-\sin^2(3x)\;\;(1).$$
If one look at a proof of the fact that $\sin^2(\color{orange}x)+\cos^2(\color{orange}x)=1$, exempli gratia here, one can deduce that the same must be true for not only $\color{orange}x$ but also for $3x$, so $\sin^2(3x)+\cos^2(3x)=1$ (look here if you still do not understand). Therefore,
$$\cos^2(3x)=1-\sin^2(3x)\;\;(2).$$
By inserting $(2)$ into $(1)$, one get that
$$1-\sin^2(3x)-\sin^2(3x)=1-2\sin(3x).$$ This means that $\cos(6x)=1-2\sin(3x)$. Therefore,
$$\sin(3x)=1-2\sin(3x).$$
Note that the correct trigonometric identity named Double-angle formula is
$$\cos 2x = 1-2\sin^2\color{red}x\implies \cos 6x = 1-2\sin^2\color{red}{3x}$$