Find exactly all solutions to the given equation

Question

Fine exactly all solutions to the give equation:

2cos^2θ = 3sinθ + 3, 0 ≤ θ ≤ 2pi

Answer 1

The Pythagorean Identity states:

sin²(x) + cos²(x) = 1

so manipulating this identity, we can get:

cos²(x) = 1 - sin²(x)

Replacing "cos²(θ)" in the equation we are given with "1 - sin²(θ)" we get:

2(1 - sin²(θ)) = 3sin(θ) + 3

2 - 2sin²(θ) = 3sin(θ) + 3

0 = 2sin²(θ) + 3sin(θ) + 1

Let w = sin(θ) then this equation becomes:

0 = 2w² + 3w + 1

0 = (2w + 1)(w + 1)

Setting each binomial equal to zero we get w = -1/2 and w = -1

Back substituting sin(θ) = w we get sin(θ) = -1/2 and sin(θ) = -1

For sin(θ) = -1/2, using the unit circle, we get θ = 7π/6 and θ = 11π/6

For sin(θ) = -1, using the unit circle, we get θ = 3π/2

Summarizing, the solutions are θ = 7π/6, θ = 11π/6, and θ = 3π/2