Posted by Kayla on Tuesday, October 16, 2007 at 11:10pm.
how do i solve this using trig identities?
2 csc^2x = 3 cot^2x-1
math - Reiny, Wednesday, October 17, 2007 at 7:22am
first change everything to sines and cosines
2/(sin^2 x) = 3(cos^2 x)/(sin^2 x) - 1
multiply by sin^2 x
2 = 3 cos^2 x - sin^2 x
but sin^2 x - 1 - cos^2 x
2 = 3 cos^2 x - 1 + cos^2 x
3 = 4cos^2 x
cos^2 x = 3/4
cos x = ±√3/2
Do you recognize the 1,√3,2 right-angled triangle? If not your calculuator should give you 30º.
if cos x = +√3/2 then x is in the first our fourth quadrants and
x = 30º or 330º
if cos x = -√3/2 then x is in the second and third quadrants and
x = 150º or 210º
(in radians the answers would be
pi/6, 11pi/6, 5pi/6,7pi/6)
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