If 0-x- 2- and -cosx-sin x- then

The correct option is D x ∈ [ #160; π/4,3π/4] given inequality gives sin x lt; cos x lt; sin x this holds for π4 lt; x lt; 3π4 ...

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Properties of Determinants

If 0≤x≤ 2π ...

Question

If $0 \leq x \leq 2 π$ and $| cos x | \leq sin x$ , then

x \in [0, \frac{π}{4}]

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x \in [\frac{π}{4}, 2 π]

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x \in [\frac{π}{4}, \frac{3 π}{4}]

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[0, π]

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Solution

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Similar questions

{x \in R : | cos x | \geq sin x} \cap [0, \frac{3 π}{2}] =

x \in (\frac{π}{4}, \frac{3 π}{4}),

\int \frac{sin x - cos x}{\sqrt{1 - sin 2 x}} e^{sin x}

cos x d x =

{sin}^{5} x - {cos}^{5} x = \frac{1}{cos x} - \frac{1}{sin x}

sin x \neq cos x)

f (x) = sin x, \forall x \in [0, \frac{π}{2}],

f (x) + f (π - x) = 2, \forall x \in (\frac{π}{2}, π]

f (x) = f (2 π - x), \forall x \in (π, 2 π],

y = f (x)

x

\sqrt{1 - sin 2 x} = sin x

x ϵ [0, π / 4]

cos x > sin x

x \in [0, \frac{π}{4}]

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