Finding the exact values of $\sin 4x - \sin 2x = 0$ So I've used the double angle formula to turn
$$\sin 4x - \sin 2x = 0$$
$$2\sin2x\cos2x - \sin2x = 0$$
$$\sin2x(2\cos2x - 1) = 0$$

$$\sin2x = 0$$
$$2x = 0$$
$$x = 0$$

$$2\cos2x - 1 = 0$$
$$2\cos2x = 1$$
$$\cos2x = \frac{1}{2}$$
$$2x = 60$$
$$x = 30$$

With this information I am able to use the unit circle to find
$$x = 0,30,180,330,360$$
However, when I looked at the answers, $$x = 0,30,90,150,180,210,270,330,360$$
Can someone tell me how to obtain the rest of the values for $x$.
Thanks
 A: Hint: $\cos(2x) = \frac{1}{2}$ has other solutions in the interval $[0, 360)$ besides $30$.  What are they?  Does $\sin(2x) = 0$ have any others besides what you find?  Ask yourself what values of $2x$ you have to look at to find all of the values of $f(x)$ for $x \in [0, 360)$.  Does this help?
An alternative (less formal) approach to this problem is to think about what $\sin(4x)$ and $\sin(2x)$ look like.  From this it is clear that there will need to be $8$ solutions in $[0, 360)$ and a little thought will tell you how they have to be spaced.  
A: Notice, you should consider all the possible values of $x$ in the respective interval, $$\sin 2x(2\cos 2x-1)=0$$
consider the following two cases, 


*

*$$\sin 2x=0$$$$\implies 2x=n\pi\ \ \ or\ \ \ x=\frac{n\pi}{2}$$
Where, $n$ is any integer


Now, for the interval $x\in [0, 2\pi]$, setting $n=0, 1, 2, 3, 4$ one should get 
$$x=\color{blue}{0, 90^\circ, 180^\circ, 270^\circ, 360^\circ}$$


*$$2\cos 2x-1\iff \cos 2x=\frac{1}{2}=\cos\frac{\pi}{3} $$ $$\implies 2x=2n\pi\pm\frac{\pi}{3}\ \ \ or\ \ \ x=n\pi\pm\frac{\pi}{6}$$


Now, for the interval $x\in [0, 2\pi]$, setting $n=0, 1, 2$ one should get 
$$x=\color{blue}{30^\circ, 150^\circ, 210^\circ, 330^\circ}$$
Hence, writing the complete solution for $x$, one should get
$$x=\color{red}{0, 30^\circ, 90^\circ, 150^\circ, 180^\circ, 210^\circ, 270^\circ, 330^\circ, 360^\circ}$$
A: Use the formula: $$\sin x-\sin y=2\sin\frac{x-y}{2}\cos\frac{x+y}{2},$$ we have:
$$\sin 4x-\sin 2x=2\sin\frac{4x-2x}{2}\cos\frac{4x+2x}{2}=2\sin x\cos 3x.$$
Now, 
$$\sin 4x-\sin 2x=0$$
$$2\sin x\cos 3x=0$$
$$\sin x\cos 3x=0$$
$$\sin x=0\Rightarrow x=k\pi, k\in\mathbb{Z},$$
and
$$\cos 3x=0\Rightarrow 3x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{3}, k\in\mathbb{Z}$$
A: $$\sin(4x)-\sin(2x)=0\Longleftrightarrow$$
$$\sin(4x)=\sin(2x)$$

Take the inverse sine of both sides:



*

*$$4x=\pi-2x+2\pi n_1\Longleftrightarrow$$
$$6x=\pi+2\pi n_1\Longleftrightarrow$$
$$x=\frac{\pi}{6}+\frac{\pi n_1}{3}$$

*$$4x=2x+2\pi n_2\Longleftrightarrow$$
$$2x=2\pi n_2\Longleftrightarrow$$
$$x=\pi n_2$$


With $n_1,n_\in\mathbb{Z}$

So we got the following solutions:
$$x=\frac{\pi n}{2}$$
$$x=\pi n\pm\frac{\pi}{6}$$
A: Using the prosthaphaeresis formula,
$$ \sin{A}-\sin{B} = 2\cos{\frac{A+B}{2}}\sin{\frac{A-B}{2}}, $$
we find
$$ 2\cos{3x}\sin{x}=0. $$
Now you just have to solve $\sin{x}=0$ or $\cos{3x}=0$. Sine has zeros at $180n$ degrees for $n$ an integer, and $\cos{y}$ has zeros at $y=180n+90$ degrees. Dividing by $3$ and checking all cases gives the result you want.
A: First you should use the natural angle unit, which is the radian. Second, you should write the general solution, which is always a congruence. 
Your equation factors as 
\begin{align*}
2\sin x\cos x(2\cos 2x -1)=0 &\iff\begin{cases}\sin x=0\\ \cos x=0 \\\cos x=\frac12
\end{cases}\\
& \iff 
\begin{cases}
x\equiv 0\mod \pi \\
x\equiv \frac\pi2\mod\pi\\2x\equiv\pm\frac\pi3\mod 2\pi\iff x\pm\frac\pi6\mod \pi
\end{cases}
\end{align*}

