How to solve $\cos(4x)=\cos\left(\frac\pi2+3x\right)$ for $0\le x\le \pi? $ Question in the title please. I know that the two cosines are equal only when their arguments (i.e what's inside the brackets) are equal, therefore it is easy to arrive at the first trivial solution being $4x=\dfrac\pi2+3x $, which gives $x=\dfrac\pi2.$
Previously when solving simple trig equations of the form, for instance, $\cos x=\dfrac12$, my strategy is as follows:
1) because what comes after the '=' is a positive constant, we are only interested in x being in the $1$st and $4$th quadrants, as cosine is only positive there. Therefore discard x in the $2$nd and $3$rd quadrants.
2) Write down general solution in the form $x=2n\pi\pm a,$ where $n=0,1,2,\dots$
As what comes after '=' in my question is not a constant, I cannot seem to adapt the above strategy.
Any help would be greatly appreciated.
 A: Well, you can actually use the following "immediate formulas"...
If $\cos x = \cos y$, then $x = \pm y + k.2\pi$
If $\sin x = \sin y$, then $x = y + k.2\pi$ or $x = (\pi - y) + k.2\pi$
If $\tan x = \tan y$, then $x = y + k.\pi$
where $k$ refers to an integer.
So for your equation, $\cos 4x = \cos (\frac{\pi}{2} + 3x)$ yields $4x = \pm (\frac{\pi}{2} + 3x) + k.2\pi$, so we get $x = \frac{\pi}{2} + k.2\pi$ or $x=-\frac{\pi}{14} + k.\frac{2\pi}{7}$. Since $0\le x \le\pi$, then the solutions for $x$ are $\frac{3\pi}{14}, \frac{\pi}{2}$, and $\frac{11\pi}{14}$. You can check that these solutions satisfy the given equation.
A: Guide:
If $2\pi\mathbb Z:=\{2\pi k\mid k\in\mathbb Z\}$ then:
$$\cos\alpha=\cos\beta\iff \alpha-\beta\in2\pi\mathbb Z\text{ or }\alpha+\beta\in2\pi\mathbb Z$$Work this out for $\alpha=4x$ and $\beta=\frac12\pi+3x$ and accept the solutions that satisfy $0\leq x\leq\pi$.
A: $$\cos x =\cos \alpha \implies x=2n\pi\pm \alpha, n=0,\pm1,\pm2,\pm3,...$$
So here $$\cos 2x = \cos (\pi/2+3x),~ x \le \pi.$$
$$\implies 2x=2n\pi \pm (\pi/2+3x) \implies 7x=2n\pi+ pi/2, x=-2m\pi+\pi/2$$ So $$x=\frac{2n\pi+\pi/2}{7}, n=1,2,3 \implies x=\frac{3\pi}{14}, \frac{\pi}{2}, \frac{11\pi}{14}.$$
