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What is root in equation cos 12x-2(sin3x)^2-1=0?

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greenbel | (Level 2) Honors

Posted September 15, 2013 at 3:27 PM via web

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What is root in equation cos 12x-2(sin3x)^2-1=0?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 15, 2013 at 3:46 PM (Answer #1)

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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` .

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