PLEASE help me on this Pre-Calculus Math problem.?

3cos^2θ - 4sinθ = 4

cos^2θ = 1 - sin^2θ

3( 1- sin^2 θ) - 4sin θ = 4

3 - 3 siv^2 θ - 4sin θ = 4

-3 sin^2 θ -4 sin θ -1 = 0

now, let x = sin θ

-3x^2 - 4 x - 1 = 0

3 x^2 + 4 x +1 = 0

now, factorize,

( 3x + 1 ) ( x + 1) =0

3x + 1=0 or x + 1 = 0

x = -1/3 or x=-1

sin θ = -1/3

θ = 3.48 or θ= 5.94

or, sin θ = -1

θ = 4.71

so,

θ = 3.48 , 4.71 , 5.94

recall that cos^2θ = 1 - sin^2θ. so you can transform the equation into the form:

3(1-sin^2θ)-4sinθ=4

3-3sin^2θ-4sinθ=4

3sin^2θ+4sinθ+1=0

now factor:

(3sinθ+1)(sinθ+1)=0

so either 3sinθ+1=0 or sinθ+1=0

consider the first case:

3sinθ+1=0

3sinθ=-1

sinθ=-1/3

θ = arcsin(-1/3) where arcsin means the inverse sine, or sin^-1

you need a calculator to evaluate this and you get θ=340.5 degrees

consider the second case:

sinθ+1=0

sin��=-1

θ = arcsin(-1) where arcsin means the inverse sine, or sin^-1

you can use a calulator or the unit circle to find that θ=270 degrees

3cos²θ - 4sinθ = 4

3(1 - sin²θ) - 4sinθ = 4

3 - 3sin²θ - 4sinθ = 4

- 3sin²θ - 4sinθ - 1 = 0

3sin²θ + 4sinθ + 1 = 0

(sinθ + 1)(3sinθ + 1) = 0

sinθ = -1

θ = arcsin(-1) = π

(3sinθ + 1) = 0

3sinθ = -1

sinθ = -1/3

θ = arcsin(-1/3) = π + 0.3398, 2π - 0.3398

θ = 3.4814, 5.9433 radians

the answer is 4π. :D Hope i helped!

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