SOLUTION: Please help me find the eXACT solutions of the given equation, in radians, that lie in the interval [0, 2π). 6 cos3(x) + 6 cos(x) = 0 I started it by factoring 6cos(x

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me find the eXACT solutions of the given equation, in radians, that lie in the interval [0, 2π). 6 cos3(x) + 6 cos(x) = 0 I started it by factoring 6cos(x      Log On


   

Question 436796: Please help me find the eXACT solutions of the given equation, in radians, that lie in the interval [0, 2π).
6 cos3(x) + 6 cos(x) = 0
I started it by factoring
6cos(x)*[cos2(x)+cos(x)]=0

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
6+cos3%28x%29+%2B+6+cos%28x%29+=+0 ==> cos3x + cosx = 0
==> cos2x cosx - sin2x sinx + cosx = 0
==> cos2x+cosx+-+2%28sinx%29%5E2%2A+cosx++%2B+cosx+=+0
==> cosx%28cos2x+-+2%28sinx%29%5E2++%2B+1%29+=+0
==> cosx%281+-+2%28sinx%29%5E2+-+2%28sinx%29%5E2+%2B+1%29+=+0
==> cosx%282+-+4%28sinx%29%5E2%29+=+0
==> cosx%281-2%28sinx%29%5E2%29+=+0
==> cosx = 0 ==> x = pi%2F2, %283pi%29%2F2;
==> Also, %28sinx%29%5E2+=+1%2F2, or sinx+=+sqrt%282%29%2F2, sinx+=+-sqrt%282%29%2F2
==> x = pi%2F4, %283pi%29%2F4, %285pi%29%2F4, %287pi%29%2F4.
==> Solution set is { pi%2F2, %283pi%29%2F2, pi%2F4, %283pi%29%2F4, %285pi%29%2F4, %287pi%29%2F4 }.