Having trouble with finding solutions to sine equation.

Question

The problem is to find solutions to $$\sin(4x -\pi/3) = \sin x$$ over the interval $[0, 2\pi ] $.

I tried expanding out $\sin(4x -\pi/3)$ but the resulting equation turns out to be very complicated. Is there any simpler way to solve this. Somehow wolfram got the exact solutions.

Answer 1

Thanks for the help (in the comments). I will attempt to find the general solution first.

I will use the fact: $$ \sin (u) = \sin(v) \implies u = v $$

Then $$ \sin(4x -\pi/3) = \sin x \\ \sin(4x -\pi/3 + 2\pi k_1) = \sin (x+ 2\pi k_2 )~~~\text{where } k_1,k_2 \in \mathbb Z \\ 4x -\pi/3 + 2\pi k_1 = x+ 2\pi k_2 \\ 3x -\pi/3 = 2\pi k_2 - 2\pi k_1 \\ 3x -\pi/3 = 2\pi( k_2 - k_1 ) \\ \qquad \qquad \qquad 3x = 2\pi( k ) + \pi/3 \quad \text{ where } k = k_2 - k_1\in \mathbb Z \\ x = \frac{2\pi }{3}k + \frac{\pi}{9}$$ And similarly $$\sin(\pi - (4x -\pi/3) ) = \sin x \\ \sin(\pi - 4x + \pi/3) ) = \sin x \\ \sin( 4/3 \pi - 4x) = \sin x \\ \sin( 4/3 \pi - 4x + 2\pi k_1) = \sin (x+ 2\pi k_2 ) \\ 4/3 \pi -4x + 2\pi k_1 = x+ 2\pi k_2 \\ 5x = 4/3 \pi + 2\pi k_1 - 2\pi k_2 \\ 5x = 4/3 \pi + 2\pi ( k_1 - k_2 ) \\ \qquad \qquad \qquad 5x = 4/3 \pi + 2\pi (k) \qquad \text{ where } k = k_2 - k_1 \in \mathbb Z \\ x = \frac{4\pi}{15} + \frac{2\pi}{5} k$$

Answer 2

$$\sin(4x-\pi/3)=\sin(x)$$ firstly, expand the first term: $$\sin(4x-\pi/3)=\sin(4x)\cos(\pi/3)-\cos(4x)\sin(\pi/3)=\frac{1}{2}\sin(4x)-\frac{\sqrt{3}}{2}\cos(4x)$$ and then you could expand all this out to get a polynomial which may be solved. Also note that if we restrict the domain of many functions the following is true:

if $f(x)=f(y)$ then $x=y$

there is an answer above which shows how to use this