Precalculus solving equations with sin and cos

Question

4(1+sin theta)=cos^2 theta

Answer 1

The given equation is

4(1+sinθ) =cos²θ..............(1)

We know

cos²θ +sin²θ =1 => cos²θ =1-sin²θ ............(2)

substituting (2) in (1), we get

4+4sinθ = 1-sin²θ

=> sin²θ +4sinθ+3=0

This is a quadratic equation in sinθ. The roots of this equation are given as

sinθ = (-4± √(4²-4*3))/2 = (-4± 2)/2 = -1,-3

The range of sinθ is [-1,1], so -3 is not possible.

So the required solution is sinθ = -1 or θ=3π/2

Answer 2

4(1 + sinθ) = 1 - sin²θ; sin²θ + 4sinθ + 3 = 0; (sinθ + 1)(sinθ + 3) = 0; sinθ = - 1; θ = 3π/2 + nπ, n = 0, ±1, ±2, ..

sinθ + 3 ≠ 0