Answers
Answered by
3
Solution :
Given, 2 cos²θ + 3 sinθ = 0
⇒ 2 (1 - sin²θ) + 3 sinθ = 0
⇒ 2 - 2 sin²θ + 3 sinθ = 0
⇒ 2 sin²θ - 3 sinθ - 2 = 0
⇒ 2 sin²θ - 4 sinθ + sinθ - 2 = 0
⇒ 2 sinθ (sinθ - 2) + 1 (sinθ - 2) = 0
⇒ (sinθ - 2) (2 sinθ + 1) = 0
⇒ 2 sinθ + 1 = 0 [ sinθ - 2 ≠ 0 ]
⇒ sinθ = - 1/2
⇒ sinθ = sin(- π/6)
∴ θ = nπ - (- 1)ⁿ . π/6 , n ∈ ℤ
Swarup1998:
:-)
Similar questions
Social Sciences,
6 months ago
Accountancy,
6 months ago
Math,
6 months ago
Geography,
1 year ago
Math,
1 year ago
Math,
1 year ago
Math,
1 year ago