I need help to see if the corresponding angles are correct

Question

$\displaystyle\cos 4x - \cos 3x=0$

$\displaystyle-2 \sin (7/2)x \cos (1/2)x= 0$

So does $\displaystyle\sin (7/2)x=\dfrac{2\pi}{7}$ and $\displaystyle\cos (1/2)x=\dfrac{\pi}{2}$?

Answer 1

You are going in the right direction though you have either miswritten or mistyped your trig identity. It should be $\cos \theta - \cos \varphi = -2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)$. So you have $\sin\left(\dfrac{7}{2}x\right) = 0 \Longrightarrow \dfrac{7}{2}x = \pi + k\pi$ and $\sin\left(\dfrac{1}{2}x\right) = 0 \Longrightarrow \dfrac{1}{2}x = \pi+k\pi$ where $k \in \Bbb{Z}$.

Answer 2

Using Prosthaphaeresis Formulas

$$\cos2x-\cos2A=1-2\sin^2x-(1-2\sin^2A)=2(\sin^2A-\sin^2x)$$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$\displaystyle\cos2x-\cos2A=2\sin(A+x)\sin(A-x)$

Now if $\displaystyle\cos2x-\cos2A=0, \sin(A+x)\sin(A-x)=0$

If $\displaystyle\sin(A+x)=0,A+x=n\pi\iff A=n\pi-x$ where $n$ is any integer

Similarly, if $\displaystyle\sin(A-x)=0,A-x=m\pi\iff A=m\pi+x$ where $m$ is any integer

$\displaystyle\implies A=r\pi\pm x\iff 2A=2r\pi\pm2x$ where $r$ is any integer