Question:
If \(2 \sin x-\cos 2 x=1\), then :
If \(2 \sin x-\cos 2 x=1\), then :
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Write cos 2x = 1 - 2 sin2x to get a quadratic in sin x, then note sin4x - sin2x = -sin2x cos2x.
Updated On: Nov 12, 2026
- \(\cos ^{4} x-\cos ^{2} x=3-\sqrt{5}\)
- \(2\left(\cos ^{4} x-\cos ^{2} x\right)=-1\)
- \(\sin ^{4} x-\sin ^{2} x=2-\sqrt{5}\)
- \(2\left(\sin ^{4} x-\sin ^{2} x\right)=1\)
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The Correct Option is C
Solution and Explanation
Use cos 2x = 1 - 2 sin2x: 2 sin x - 1 + 2 sin2x = 1, so sin2x + sin x - 1 = 0.
Hence sin2x = 1 - sin x, and sin x = (-1 + √5)/2 (the other root is below -1).
Then cos2x = 1 - sin2x = sin x = (√5 - 1)/2.
sin4x - sin2x = -sin2x cos2x = -(1 - sin x)(sin x) = -(sin x - sin2x) = -(2 sin x - 1) = -(√5 - 2) = 2 - √5.
For option (1): cos4x - cos2x = -sin2x cos2x = 2 - √5, not 3 - √5, so (1), (2), (4) are false.
Answer: (3) \(\sin ^{4} x-\sin ^{2} x=2-\sqrt{5}\)
Hence sin2x = 1 - sin x, and sin x = (-1 + √5)/2 (the other root is below -1).
Then cos2x = 1 - sin2x = sin x = (√5 - 1)/2.
sin4x - sin2x = -sin2x cos2x = -(1 - sin x)(sin x) = -(sin x - sin2x) = -(2 sin x - 1) = -(√5 - 2) = 2 - √5.
For option (1): cos4x - cos2x = -sin2x cos2x = 2 - √5, not 3 - √5, so (1), (2), (4) are false.
Answer: (3) \(\sin ^{4} x-\sin ^{2} x=2-\sqrt{5}\)
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