2sin2θ + 3cosθ - 3 = 0 হলে, θ এর মান কত? যেখানে θ সূক্ষ্মকোণ
Solution
Correct Answer: Option A
2sin²θ + 3cos θ = 0
[since sin²θ=1- cos²θ]
বা, 2(1- cos²θ) + 3cosθ - 3= 0
বা, 2 - 2cos²θ + 3cosθ - 3= 0
বা, -2cos²θ + 3cosθ - 1 = 0
বা, - (2cos²θ - 3cosθ + 1) = 0
বা, 2cos²θ - 3cosθ + 1 = 0
বা, 2cos²θ - 2cosθ - cosθ + 1 = 0
বা, 2cosθ(cosθ - 1)- (cosθ - 1) = 0
∴ (cosθ - 1)(2cosθ - 1) = 0
এখন,
2cosθ - 1 = 0
বা, cosθ = 1/2
বা, cosθ =cos60°
∴ θ =60°
অথবা
বা, cosθ - 1 = 0
বা, cosθ = 1
বা, cosθ = Cos0°
∴ θ = 0° [গ্রহণযোগ্য নয়]
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