Question

The number of solutions of the pair of equations 2 $si{n}^2$θ - $co{s}^2$θ = 0 and 2 $si{n}^2$θ - 3 sin θ = 0, in the interval [0, 2π] is

• A. Zero
• B. One
• C. Two
• D. Four

Answer 1

The correct option is C

Two

We want to find the values of θ which will satisfy both the equation.

2 $si{n}^2$θ - $co{s}^2$θ = 0

⇒ 2 $si{n}^2$θ - (1- 2 $si{n}^2$θ) = 0

⇒ 4 $si{n}^2$θ - 1= 0
⇒ $si{n}^2$θ = $\frac{1}{4}$
⇒ θ = nπ± $\frac{\pi }{6}$ ..........(1)

[if $si{n}^2$θ = $si{n}^2$α, θ = nπ±α]
2 $si{n}^2$θ - 3 sin θ = 0

⇒ we can form a quadratic in sin θ and solve this.

2(1- $si{n}^2$θ) - 3 sinθ = 0

⇒ 2 - 2 $si{n}^2$θ - 3 sin θ = 0

⇒ 2 $si{n}^2$θ + 3 sin θ - 2 = 0

⇒ 2 $si{n}^2$θ - sin θ + 4 sin θ - 2 = 0
⇒ 2 sin θ (sin θ - $\frac{1}{2}$) + 4 (sinθ - $\frac{1}{2}$) = 0
⇒ (2 sin θ + 4) (sin θ – $\frac{1}{2}$) = 0

⇒ sin θ = $\frac{1}{2}$ [sin θ ≠ –2]

⇒ θ = nπ + (–1)ⁿ $\frac{1}{2}$ . . . . . (2)

We can substitute different values of n (n = 0, 1, 2, 3 . . . . .) in (1) and (2), then pick the common values. If we do that we will get $\frac{\pi }{6}$ and $\frac{5\pi }{6}$ as common values

The first equation gave us sin² θ = $\frac{1}{4}$ or sin θ = ∓ $\frac{1}{2}$ and second equation is sin θ = $\frac{1}{2}$. The common part is sin θ = $\frac{1}{2}$. This has two solutions in [0, 2π]. They are $\frac{\pi }{6}$ and $\frac{5\pi }{6}$.

⇒ two values