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

so

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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