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What is root in equation cos 12x-2(sin3x)^2-1=0?
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You should use the double angle identity and the half angle identity, such that:
`cos 12x = cos 2*(6x) = 2cos^2 6x - 1`
`sin^2 (3x) = sin^2 ((6x)/2) = (1 - cos 6x)/2`
Replacing ` 2cos^2 6x - 1` for `cos 12x` and `(1 - cos 6x)/2` for `sin^2 (3x)` , yields:
`2cos^2 6x - 1 - 2*(1 - cos 6x)/2 - 1 = 0`
Reducing duplicate factors yields:
`2cos^2 6x - 1 - 1 + cos 6x - 1 = 0`
You need to arrange the terms so the powers of cosines are in descending order, such that:
`2cos^2 6x + cos 6x - 3 = 0`
You should come up with the following substitution, such that:
`cos 6x = y`
`2y^2 + y - 3 = 0`
Using quadratic formula, yields:
`y_(1,2) = (-1+-sqrt(1 + 24))/4`
`y_(1,2) = (-1+-5)/4 => y_1 = 1 ; y_2 = -3/2`
You need to solve for x the following equations, such that:
`cos 6x = 1 => 6x = +-cos^(-1) 1 + 2n*pi`
`6x = 2n*pi => x = (2n*pi)/6 => x = (n*pi)/3`
`cos 6x =-3/2` is invalid hence `cos 6x` cannot be smaller than -1.
Hence, evaluating the general solution to the given trigonometric equation, yields `x = (n*pi)/3` .
Posted by sciencesolve on September 15, 2013 at 3:46 PM (Answer #1)
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