Question

please help me to do it​

Answer 1

Let's take theta = A for the sake of convenience.

So, we have :-

=> sinA + sin7A = sin4A

=> 2sin[(A+7A)/2].cos[(A-7A)/2] = sin4A

=> 2sin4A.cos3A = sin4A

=> 2sin4A.cos3A - sin4A = 0

=> sin4A(2cos3A - 1) = 0

Therefore,

sin4A = 0 or 2cos3A - 1 = 0

4A = nπ or cos3A = ½

A = nπ/4 or 3A = 2nπ ± π/3

A = nπ/4 or A = (6n ± 1)π/9

Here, n € Z

Hope, it'll help you.....