Solve the equation sin² x - sin x - 2 = 0, for x on the interval 0 ≤ x < 2π.

Concept:

The solution to the quadratic equation ax2 + bx + c = 0 is given by: \(\rm x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

• sin (-θ) = -sin θ.
• sin θ = sin (2nπ + θ).

Calculation:

sin2 x - sin x - 2 = 0

Using the formula for the roots of a quadratic equation: 

⇒ \(\rm \sin x = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-2)}}{2(1)}\)

⇒ \(\rm \sin x = \dfrac{1 +\sqrt{9}}{2}\) OR \(\rm \sin x = \dfrac{1 -\sqrt{9}}{2}\)

⇒ sin x = 2 OR sin x = -1.

∵ -1 ≤ sin x ≤ 1, ∴ sin x = 2 is not possible.

sin x = -1 = \(\rm -\sin\dfrac{\pi}{2}=\sin\left(-\dfrac{\pi}{2}\right)=\sin\left(2n\pi-\dfrac{\pi}{2}\right)\), n ∈ Z.

For n = 0, x = \(-\dfrac{\pi}{2}\).

For n = 1, x = \(\dfrac{3\pi}{2}\).

For n = 2, x = \(\dfrac{7\pi}{2}\).

The only value of x on the interval 0 ≤ x < 2π is x = \(\dfrac{3\pi}{2}\).

Additional Information

• Period of sin θ is 2π. ⇒ sin θ = sin (2nπ + θ).
• Period of cos θ is 2π. ⇒ cos θ = cos (2nπ + θ).
• Period of tan θ is π. ⇒ tan θ = tan (nπ + θ).