Let - -3 then the solution set of the inequality log sin- 2-cos 2x - log sin- 1-sin x where x - 0-2- and x-2 is

The correct option is D ( 0,π/2 )∪ ( π/2,π ) log _sinα #160; #160; ( 2 cos ^ 2 x #160; ) #160; log _sinα #160; #160; ( 1 sin x # ...

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Let α =π/3 ...

Question

Let $α = \frac{π}{3}$ then the solution set of the inequality ${log}_{sin α} (2 - {cos}^{2} x) < {log}_{sin α} (1 - sin x)$ where $x ϵ (0, 2 π)$ and $x \neq \frac{π}{2}$ is

(α, π - α)

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(α, \frac{π}{2}) \cup (\frac{π}{2}, π - α)

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(\frac{π}{3}, \frac{5 π}{6})

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(0, \frac{π}{2}) \cup (\frac{π}{2}, π)

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Solution

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Similar questions

α = \frac{π}{3},

{log}_{sin α} (2 - {cos}^{2} x) < {log}_{sin α} (1 - sin x),

x ε (0.2 π)

x \neq \frac{π}{2},

\int_{0}^{1} {tan}^{- 1} (\frac{tan x}{2}) d x = α

\int_{0}^{1} {tan}^{- 1} (\frac{tan x - 2 cot x}{3}) d x

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(- 1)^{n}$ α where α ∈ [ - $\frac{π}{2}$ , $\frac{π}{2}$ ] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{π}{2}$ , $\frac{π}{2}$ ) n ∈ I

f : [0, \frac{π}{2}] \to [0, 1]

f (0) = 0, f (\frac{π}{2}) = 1.

[0, \frac{π}{2}] \to [0, 1]

f (0) = 0, f (\frac{π}{2}) = 1,

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