SOLUTION: How do you solve: sin^2(x)+cos(2x)-cos(x)=0 I know that cos(2x) can be equal to cos^2(x)-sin^2(x), or it can be equal to 2cos^2(x)-1, or it can be equal to 1-2sin^2(x) My

Algebra ->  Trigonometry-basics -> SOLUTION: How do you solve: sin^2(x)+cos(2x)-cos(x)=0 I know that cos(2x) can be equal to cos^2(x)-sin^2(x), or it can be equal to 2cos^2(x)-1, or it can be equal to 1-2sin^2(x) My      Log On


   



Question 454350: How do you solve:
sin^2(x)+cos(2x)-cos(x)=0
I know that cos(2x) can be equal to cos^2(x)-sin^2(x), or it can be equal to 2cos^2(x)-1, or it can be equal to 1-2sin^2(x)
My math teacher told us to use the formula that has the trig-word in it that is in the equation, but since both sin and cos are in the equation, I'm not sure what to use.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
if you use cos^2(x)-sin^2(x) then you will have sin^2(x)+cos^2(x)-sin^2(x)
which eliminates sin completely.
leaving you cos^(2x)-cos(x)