Expert: Sherman D. - 10/16/2008

Question
2cos^2(t)=3sin(t)
5tan^2(t)+3=9sec(t)

Answer
2cos(t)^2 = 3sin(t)
2(1 - sin(t)^2) = 3sin(t)
2 - 2sin(t)^2 = 3sin(t)
2sin(t)^2 + 3sin(t) - 2 = 0

Using the quadratic formula

sin(t) = (-3 ± sqrt((3^2 - 4(2)(-2)))/(2(2))
sin(t) = (-3 ± sqrt(9 + 16))/4
sin(t) = (-3 ± sqrt(25))/4
sin(t) = (-3 ± 5)/4
sin(t) = (-8/4) or (2/4)
sin(t) = -2 or (1/2)
since we can't get sin^-1(-2)
sin(t) = (1/2)
t = 30° and 150°

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5tan(t)^2 + 3 = 9sec(t)
5(sec(t)^2 - 1) + 3 = 9sec(t)
5sec(t)^2 - 5 + 3 = 9sec(t)
5sec(t)^2 - 9sec(t) - 2 = 0

using the quadratic formula

sec(t) = (-(-9) ± sqrt((-9)^2 - 4(5)(-2)))/(2(5))
sec(t) = (9 ± sqrt(81 + 40))/10
sec(t) = (9 ± sqrt(121))/10
sec(t) = (9 ± 11)/10
sec(t) = (20/10) or (-2/10)
sec(t) = 2 or (-1/5)
1/cos(t) = 2 or (-1/5)
cos(t) = (1/2) or -5
since we can't get cos^-1(-5)
cos(t) = (1/2)
t = 60° and 300°