SOLUTION: for the interval of 0-2(pi) solve the equation {{{cos4xcos2x+sin4xsin2x=sqrt(3)/2

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Question 808770: for the interval of 0-2(pi) solve the equation for+the+interval+of+0-2%28pi%29+solve+the+equation+cos4xcos2x+sin4xsin2x=sqrt(3)/2
Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
(1) cos(4x)cos(2x) + sin(4x)sin(2x) = sqrt(3)/2
Use the difference formula
(2) cos(a-b) = cos(a)cos(b) + sin(a)sin(b) to obtain
(3) cos(4x - 2x) = cos(4x)cos(2x) + sin(4x)sin(2x)
Now equate (1) and (3) to get
(4) cos(4x - 2x) = sqrt(3)/2 or
(5) cos(2x) = sqrt(3)/2
Use your calculator to evaluate
(6) 2x = arccos(sqrt(3)/2) and get
(7) 2x = 30 or
(8) x = 15 degrees
This is also the solution at
(9) x = 15+180 or
(10) x = 195 degrees
Answer: x = {15,195} degrees