Question

Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

sin⁴x = sin²x

Answer 1

sin⁴x – sin²x = 0

sin² x (sin² x – 1) = 0

sin² x [– (1 – sin² x)] = 0

sin²x × – cos²x = 0

– sin²x cos²x = 0

(sin x cos x)² = 0

`(1/2 xx 2 sin cos x)^2` = 0

`1/4 sin 2x` = 0

sin 2x = 0

The general solution is

2x = nπ, n ∈ Z

x = `("n"pi)/2`, n ∈ Z

When n = 0, x = `(0 xx pi)/2` = 0 ∉ (0°, 360°)

When n = 1, x = `pi/2` = ∈ (0°, 360°)

When n = 2, x = `(2pi)/2` = π ∈ (0°, 360°)

When n = 3, x = `(3pi)/2` = ∈ (0°, 360°)

When n = 4, x = `(4pi)/2` = 2π ∉ (0°, 360°)

∴ The values of x are `pi/2`, π, `(3pi)/2`