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how do i solve this using trig identities?

2 csc^2x = 3 cot^2x-1

  • math -

    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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