Precalculus solving equations of sin and cos

Question

Solve the equations.

a) sin(2 theta)+sin(4 theta)=0

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

Answer 1

a)

Use the formula

sin(x) + sin(y) = 2 sin((x+y)/2) cos((x-y)/2)

sin(2θ) + sin(4θ) = 0

2 sin(3θ) cos(θ) = 0

This can happen only if one of the factors is 0.

Therefore we get two sets of solutions.

1)

sin(3θ) = 0

3θ = kπ, for any integer k

θ = kπ/3, for any integer k

2)

cos(θ) = 0

θ = π/2 + kπ, for any integer k

b)

4(1 + sin(θ)) = cos²(θ)

4 + 4sin(θ) = 1 - sin²(θ)

sin²(θ) + 4 sin(θ) + 3 = 0

This is a quadratic equation with two solutions

1) sin(θ) = (-4 + √(4² - 12)) / 2 = -1

θ = 2kπ - π/2 , for any integer k

2) sin(θ) = (-4 - √(4² - 12)) / 2 = -3

There is no θ satisfying that.