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

Given that: \[ sin 2 x - sin x - 2 = 0 \] \[ (sin x - 2)(sin x + 1) = 0 \] \[ sin x = -1 \] Hence: \[ x = 3π/2 \]

Solve the equation sin2 x... - Solved

2 x - sin x - 2 = 0, for x on the interval 0 ≤ x..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> 2 x - sin x - 2 = 0, for x on the interval 0 ≤ x..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> 2 x - sin x - 2 = 0, for x on the interval 0 ≤ x..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />

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Question 38

Mathematics Trigonometric Equations Hard

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

(A) \(\dfrac{3\pi}{2}\)

(B) \(\dfrac{\pi}{4},\dfrac{2\pi}{7}\)

(C) \(\dfrac{2\pi}{3},\dfrac{2\pi}{5}\)

(D) None of these

✓ Correct Answer: Option A Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Given that: \[ \sin^2 x - \sin x - 2 = 0 \] \[ \Rightarrow (\sin x - 2)(\sin x + 1) = 0 \] \[ \Rightarrow \sin x = -1 \] Hence: \[ x = \frac{3\pi}{2} \]

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