Anyone that can help me get started on this equation, $$\sin(4x)=\cos(3x).$$ I tried $$\sin(4x)=2\sin(2x)\cos(2x)=4\sin(x)\cos(x)\bigl(2\cos^2(x)-1\bigr)$$ and $$\cos(3x)=4\cos^3(x)-3\cos(x)$$ to get $$4\sin(x)\cos(x)\bigl(2\cos^2(x)-1\bigr)=4\cos^3(x)-3\cos(x)$$ and then square both sides to get $\sin^2(x)=1-\cos^2(x)$ but the polynomial equation does not seem to be an easy one.
Any better method? TIA.
On
$$\sin{(4x)}=\sin{\left(\frac{\pi}2-3x\right)}$$ $$\therefore 4x=2n \pi + \frac {\pi} {2} -3x\text{ where }n \in \mathbb {Z}$$
Now you can solve it.
On
Use the fact that $\cos(x) = \sin\left(\frac{\pi}{2}-x\right)$ to your advantage.
Setting this up, we get:
$$\sin(4x) = \sin\left(\frac{\pi}{2}-3x\right) \implies 4x = \frac{\pi}{2} \pm 3x + 2\pi k$$
where $k$ is an arbitrary integer. The $\pm$ comes from the fact that $\sin(x) = \sin(\pi - x)$ which is the other angle that shares the same $y$ value on the unit circle.
On
Since $\sin x=\cos (\pi/2-x)$ and $\cos x=\cos y\iff x-y=2k\pi$ or $x+y=2k\pi$ for some $k\in\Bbb Z$ we have: $$\sin (4x)=\cos (\frac{\pi}2-4x)=\cos (3x)\iff \frac{\pi}2-x=2k\pi\lor \frac{\pi}2-7x=2k\pi.$$ In first case we get $x=\pi/2 -2k\pi$, in the second $x=\pi/14-k\pi/7$, for all $k\in\Bbb Z$.
The simplest way to solve this equation is to transform into this equivalent form: \begin{align}\cos\Bigl(\frac\pi2-4x\Bigr)=\cos 3x&\iff \frac\pi2-4x\equiv \pm 3x\mod 2\pi\iff \frac\pi 2\equiv\begin{cases}7x,\\[-1ex]\text{or}\\[-2ex]x \end{cases}\bmod 2\pi \\ &\iff \begin{cases}x\equiv \frac{\pi}{14}\mod\frac{2\pi}7,\\[1ex] x\equiv\frac\pi 2\mod 2\pi. \end{cases} \end{align}