SOLUTION: Solve for x between 0 and 2pi (both included):
sin(4x) = cos(2x).
I changed sin(4x) into sin(2x+2x) so that I could use the identity so it would then become sin 2xcos2x + cos2xs
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-> SOLUTION: Solve for x between 0 and 2pi (both included):
sin(4x) = cos(2x).
I changed sin(4x) into sin(2x+2x) so that I could use the identity so it would then become sin 2xcos2x + cos2xs
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Question 1064659: Solve for x between 0 and 2pi (both included):
sin(4x) = cos(2x).
I changed sin(4x) into sin(2x+2x) so that I could use the identity so it would then become sin 2xcos2x + cos2xsin2x = cos 2x. I collected the like terms to give: 2sin 2xcos2x = cos 2x. My question is if it's possible to divide both sides by cos 2x? Or would I lose some of the values by doing so? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! I collected the like terms to give: 2sin 2xcos2x = cos 2x. My question is if it's possible to divide both sides by cos 2x? Or would I lose some of the values by doing so?
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don't divide by it.
2sin 2xcos2x - cos 2x = 0
cos(2x)*(2sin(2x) - 1) = 0
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cos(2x) = 0
sin(2x) = 1/2
etc
