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Solve for `theta` and show fundamental/general solution: sec^4θ - 2sec^2θ -...
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You need to remember that secant function is rational function such that: `sec^4theta = 1/(cos^4theta) = (1 + tan^2 theta)^2`
`sec^2 theta = 1/(cos^2 theta) = 1 + tan^2 theta`
Hence, you need to write the equation in terms of tangent such that:
`(1 + tan^2 theta)^2 - 2(1 + tan^2 theta) - 4tan^2 theta = -4`
You need to expand the binomial such that:
`1 + 2tan^2 theta + tan^4 theta - 2 - 2tan^2 theta - 4tan^2 theta = -4`
Reducing like terms yields:
`tan^4 theta - 4tan^2 theta + 3 = 0`
You should come up with the substitution `tan^2 theta = x` such that:
`x^2 - 4x + 3 = 0`
You need to use quadratic formula to find the solutions to equation such that:
`x_(1,2) = (4+-sqrt(16 - 12))/2 =gt x_(1,2) = (4+-2)/2`
`x_1 = 3 ; x_2 = 1`
You need to find `theta` , hence `tan^2 theta = x_1 =gt tan^2 theta = 3 =gt tan theta = +- sqrt 3 =gt theta = +-pi/3 + n*pi`
`tan^2 theta = x_2 =gt tan^2 theta= 1=gt tan theta = +- 1 =gt theta = +-pi/4 + n*pi`
Hence, the general solutions to trigonometric equation are `theta = +-pi/4 + n*pi ` and `theta = +-pi/3 + n*pi.`
Posted by sciencesolve on February 16, 2012 at 12:21 AM (Answer #1)
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