The solution set of ∣ sin x ∣ ≤ ∣ cos 2 x ∣ contains

Squaring gives cos 2 2 x ≥ sin 2 x , or ( 1 + cos 2 x ) ( 2 cos 2 x − 1 ) ≥ 0 . Thus cos 2 x = − 1 (isolated points x = nπ + π /2 ) or cos 2 x ≥ 1/2 (the closed intervals in A). The sets in A, B and C are contained in the solution set; D is not.

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MathematicsTrigonometric Ratios & IdentitiesBasic Identities & T RatiosMedium2 minQB

MathematicsMediummultiple choice

The solution set of $∣ sin x ∣ \leq ∣ cos 2 x ∣$ contains

Options:(select one or more)

Answer:

A, B, C

Solution:

Squaring gives $cos^{2} 2 x \geq sin^{2} x$ , or $(1 + cos 2 x) (2 cos 2 x - 1) \geq 0$ . Thus $cos 2 x = - 1$ (isolated points $x = nπ + π /2$ ) or $cos 2 x \geq 1/2$ (the closed intervals in A). The sets in A, B and C are contained in the solution set; D is not.

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Basic Identities & T Ratios

⏱ 2mℹ️ Source: QB

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