Trigonometry

Solve the equation for solutions in the interval 0<=theta<2pi
Problem 1.
3cot^2-4csc=1

My attempt:

3(cos^2/sin^2)-4/sin=1
3(cos^2/sin^2) - 4sin/sin^2 = 1
3cos^2 -4sin =sin^2
3cos^2-(1-cos^2) =4sin
4cos^2 -1 =4sin
Cos^2 - sin=1/4
(1-sin^2) - sin =1/4
-Sin^2 - sin =-3/4
Sin(sin+1) =3/4
Sin= 3/4. Or sin =-1/4
I don't think these give the right answers for theta. I keep getting an error on my calculator

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  1. from
    3cos^2 Ø -4sinØ =sin^2 Ø , why not replace the cos^2 Ø

    3(1 - sin^2 Ø) - 4sinØ - sin^2 Ø = 0
    ..
    4sin^2 Ø + 4sinØ - 3 = 0

    (2sinØ - 1)(2sinØ + 3) = 0
    sinØ = 1/2 or sinØ = -3/2 which is not possible

    sinØ = 1/2
    Ø = π/6 or 5π/6 , (30° or 150°)

    your solution breaks down here:
    Sin(sin+1) =3/4
    Sin= 3/4. Or sin =-1/4
    you can only set factors equal to zero if the right side of your equation is zero.

    My second concern is your poor usage of trig function names.
    You need an argument after the trig name
    e.g. sinØ = .3
    sin = .3 is meaningless, and is just that, a sin .
    that's like saying √ = 2

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  2. You cannot work this:
    sin(sin+1) = 3/4
    if sin=3/4, sin+1 = 7/4
    This is why we always set things to zero. If a product is zero, one of the factors nust be zero.
    sin^2+sin = 3/4
    4sin^2+4sin-3=0
    (2sin-1)(2sin+3) = 0
    sin = 1/2 or sin = -3/2
    only sin = 1/2 is allowed, so
    θ = π/6, 5π/6

    or, since cot^2 = csc^2-1, you have

    3(csc^2-1)-4csc - 1 = 0
    3csc^2 - 4csc - 4 = 0
    (3csc+2)(csc-2) = 0
    csc = 2 or -2/3
    |csc| is never less than 1, so
    csc = 2 is the only solution.
    Same as above

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