# 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 ->  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

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 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(9135)   (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)