Question

In 2cos2θ + 3sinθ = 0, then the general value of θ is

Answer 1

Answer:

2(1−sin2θ)+3sinθ=0. 2−2 sin2θ+3sinθ=0. 0=2sin2θ−3sinθ−2. 0=2sin2 θ−4sinθ+sinθ−2. 0=2sinθ(sinθ−2)+1(sinθ− 2).

Answer 2

Given,

→ 2 cos²θ + 3 sinθ = 0

Since cos²θ = 1 - sin²θ,

→ 2 (1 - sin²θ + 3 sinθ = 0

→ 2 - 2 sin²θ + 3 sinθ = 0

Multiplying by -1,

→ 2 sin²θ - 3 sinθ - 2 = 0

→ 2 sin²θ - 4 sinθ + sinθ - 2 = 0

→ 2 sinθ (sinθ - 2) + (sinθ - 2) = 0

→ (2 sinθ + 1)(sinθ - 2) = 0

Since sinθ can't be 2,

→ 2 sinθ + 1 = 0

→ sinθ = -1/2

⇒ θ = nπ + (-1)ⁿ · (- π / 6)

where n ∈ Z.