`sin 3 alpha = 4 sin alpha sin(x + alpha) sin(x-alpha)`

Options

A. `n pi pm pi//4, AA n in Z`
B. `n pi pm pi//3, AA n in Z`
C. `n pi pm pi//9, AA n in Z`
D. `n pi pm pi//12, AA n in Z`

Answer

B

Explanation

We have `sin 3 alpha =4 sin alpha (sin^(2) x-sin^(2) alpha)`
or `3 sin alpha -4 sin^(3) alpha =4 sin alpha sin^(2) x-4 sin^(3) alpha`
or `3 sin alpha =4 sin alpha sin^(2)x`
If `sin alpha ne0, sin^(2) x=3//4=(sqrt(3)//2)^(2)= sin^(2) (pi//3)`,
Therefore `x= n pi pm pi//3, AA n in Z`
If `sin alpha =0`, i.e., `alpha =n pi`, equation becomes an identity.