SOLUTION: Find the solutions which are in the interval[0,2π) of sin5x-sinx=2cos3x

Algebra ->  Trigonometry-basics -> SOLUTION: Find the solutions which are in the interval[0,2π) of sin5x-sinx=2cos3x       Log On


   

Question 371825: Find the solutions which are in the interval[0,2π) of sin5x-sinx=2cos3x
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
sin5x-sinx=2cos3x, same as 2cos3xsin2x = 2cos3x, from the identity
sinA+-+sinB+=+2cos%28%28A%2BB%29%2F2%29sin%28%28A-B%29%2F2%29.
This means, after transposition and factoring, cos3x(sin2x - 1) = 0.
Thus cos3x = 0 or sin2x = 1,
3x = pi%2F2 or 3pi%2F2, or 2x = pi%2F2.
x = pi%2F6 or pi%2F2, or x = pi%2F4.