How to find all possible x's with an interval of $[0, 2 \pi/7]$ when given the equation of $2 \sin(7x)-1=0$? So, as the title says, I'm having trouble finding the possible values of x in the following equation with an interval set to [0, 2pi/7]:
2sin(7x)-1=0
I'm a freshman in university and this was not explained during the lecture and during the study group, the assistant only went over it briefly for about 10 minutes without really explaining what she was writing on the board. Other students who did understand said she got way off topic and her method was extremely advanced, which explains why I was completely lost.
If anyone could help, it would be greatly appreciated.
 A: Start with a graph, so you get insight:

A: With the value $\frac12$ of the sine you should associate almost automatically the angle $\frac\pi6$ and by reflection on the vertical axis also $\pi-\frac\pi6=\frac{5\pi}6$ (as $\frac{\sqrt{k}}2$, $k=0,1,2,3,4$ are the sines of $0,\frac\pi6,\frac\pi4,\frac\pi3,\frac\pi2$).
These angles and their equivalents by full $2\pi$ rotations now have to be set equal to $7x$,
$$
7x=\frac\pi6+2k\pi~~\text{ or }~~7x=\frac{5\pi}6+2k\pi.
$$
A: Your equation is equivalent to $\sin 7x=1/2=\sin(π/6),$ or $$\sin 7x-\sin(π/6)=0.$$ Change the LHS into a product using $\sin a-\sin b=2\cos(a+b)/2\cos(a-b)/2.$ Then you have $$2\cos(42x+π)/12\cos(42x-π)/12=0.$$ Thus, we have that $\cos(42x+π)/12=0,$ or $\cos(42x-π)/12=0.$ I'll do only the first and leave the second for you since they're similar. The first implies $$\frac{42x+π}{12}=\frac{(2k-1)π}{2},$$ where $k$ is any integer. Thus $$x=\frac{12πk-7π}{42}.$$ But we want $x$ to be in the interval $[0,3π/7].$ Thus we must find all $k$ so that $$0\le \frac{12πk-7π}{42}\le 3π/7.$$ This simplifies to $0\le 12k\le 25.$ Thus, we have $k=0,1,2.$ Thus, three solutions are $$x=\frac{12πk-7π}{42},$$ with $k=0,1,2.$ To find the others, solve the second equation above similarly.
