Question:
Statement-1: The number of common solutions of the trigonometric equations \(2\sin^2\theta - \cos 2\theta = 0\) and \(2\cos^2\theta - 3\sin\theta = 0\) in the interval \([0, 2\pi]\) is two.
Statement-2: The number of solutions of the equation, \(2\cos^2\theta - 3\sin\theta = 0\) in the interval \([0, \pi]\) is two.
Statement-1: The number of common solutions of the trigonometric equations \(2\sin^2\theta - \cos 2\theta = 0\) and \(2\cos^2\theta - 3\sin\theta = 0\) in the interval \([0, 2\pi]\) is two.
Statement-2: The number of solutions of the equation, \(2\cos^2\theta - 3\sin\theta = 0\) in the interval \([0, \pi]\) is two.
Statement-2: The number of solutions of the equation, \(2\cos^2\theta - 3\sin\theta = 0\) in the interval \([0, \pi]\) is two.
Show Hint
Solve each equation as a quadratic in \(\sin\theta\) and count the solutions in the given interval.
Updated On: Oct 31, 2026
- Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1.
- Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
- Statement-1 is false; Statement-2 is true.
- Statement-1 is true; Statement-2 is false.
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The Correct Option is B
Solution and Explanation
First equation: \(2\sin^2\theta - (1 - 2\sin^2\theta) = 0\), so \(4\sin^2\theta = 1\) and \(\sin\theta = \pm\tfrac12\).
Second equation: \(2(1 - \sin^2\theta) - 3\sin\theta = 0\), that is \(2\sin^2\theta + 3\sin\theta - 2 = 0\), so \((2\sin\theta - 1)(\sin\theta + 2) = 0\). Since \(\sin\theta \ne -2\), we get \(\sin\theta = \tfrac12\).
Common solutions: \(\sin\theta = \tfrac12\) in \([0, 2\pi]\) gives \(\theta = \pi/6,\ 5\pi/6\). So there are 2 and Statement-1 is true.
Statement-2: \(\sin\theta = \tfrac12\) in \([0, \pi]\) gives \(\pi/6\) and \(5\pi/6\), two solutions, so Statement-2 is true.
Statement-2 counts solutions in a different interval and does not explain why the two equations share exactly those solutions (that comes from the factorisation and from \(\sin\theta = \tfrac12\) satisfying the first equation), so it is not a correct explanation.
Answer: (2) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
Second equation: \(2(1 - \sin^2\theta) - 3\sin\theta = 0\), that is \(2\sin^2\theta + 3\sin\theta - 2 = 0\), so \((2\sin\theta - 1)(\sin\theta + 2) = 0\). Since \(\sin\theta \ne -2\), we get \(\sin\theta = \tfrac12\).
Common solutions: \(\sin\theta = \tfrac12\) in \([0, 2\pi]\) gives \(\theta = \pi/6,\ 5\pi/6\). So there are 2 and Statement-1 is true.
Statement-2: \(\sin\theta = \tfrac12\) in \([0, \pi]\) gives \(\pi/6\) and \(5\pi/6\), two solutions, so Statement-2 is true.
Statement-2 counts solutions in a different interval and does not explain why the two equations share exactly those solutions (that comes from the factorisation and from \(\sin\theta = \tfrac12\) satisfying the first equation), so it is not a correct explanation.
Answer: (2) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.
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