You can put this solution on YOUR website! sin(4x) + sin(x) = 0
Rewrite the equation:
4x = (sin^−1)(sin(x)). The right hand side is either x or π−x, or either of these plus any integer multiple of 2π. So, in the following equation find all values of n that give results between 0 and π:
4x = x + 2nπ
or
4x=π−x + 2nπ
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example:
x = (2nπ)/3 or x = ((2n+1)π)/5
