Question

Solve the following equations:
2 cos²θ + 3 sin θ – 3 = θ

Answer 1

2 cos²θ + 3 sin θ – 3 = θ

2(1 – sin²θ)+ 3 sin θ – 3 = θ

2 – 2 sin²θ + 3 sin θ – 3 = θ

– 2 sin²θ + 3 sin θ – 1 = θ

2 sin² θ – 3 sin θ + 1 = θ

2 sin²θ – 2 sin θ – sin θ + 1 = θ

2 sin θ (sin θ – 1) – (sin θ – 1) = θ

(2 sin θ – 1)(sin θ – 1) = 0

2 sin θ – 1 = 0 or sin θ – 1 = θ

sin θ = `1/2` or sin θ = 1

To find the general solution of’ sin θ = `1/2`

sin θ = `1/2`

sin θ = `sin  pi/6`

The general solution is θ = `"n"pi + (- 1)^"n"  pi/6`, n ∈ Z

To find the general solution of sin θ = 1

sin θ = 1

sin θ = `pii/2`

The general solution is θ = `"n"pi + (- 1)^"n"  pi/6`, n ∈ Z

∴ The required solutions are

θ = `"n"pi + (- 1)^"n"  pi/6`, n ∈ Z

or

θ = `"n"pi + (- 1)^"n"  pi/6`, n ∈ Z