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Determine the general solution of : `cos^2 theta + 3sin theta = - 3` Thank you
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You need to use the basic trigonometric formula `1 - sin^2 theta = cos^2 theta` , to write the equation in terms of `sin theta` , only, such that:
`(1 - sin^2 theta) + 3sin theta = -3`
`1 - sin^2 theta + 3sin theta + 3 = 0`
`- sin^2 theta + 3sin theta +4 = 0`
Factoring out -1 yields:
`-1(sin^2 theta - 3sin theta - 4) = 0`
Replacing t for `sin theta` yields:
`t^2 - 3t - 4 = 0`
You may use quadratic formula to evaluate `t_(1,2)` such that:
`t_(1,2) = (3+-sqrt(9 + 16))/2 => t_(1,2) = (3+-5)/2`
`t_1 = 4, t_2 = -1`
Replacing back sin theta for t yields:
`sin theta = 4` invalid since `sin theta <= 1`
`sin theta = -1 => theta = (-1)^(n+1)(pi/2) + n*pi`
Hence, evaluating the general solution to the given equation, yields `theta = (-1)^(n+1)(pi/2) + n*pi.`
Posted by sciencesolve on June 3, 2013 at 3:24 PM (Answer #1)
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