The solution set of inequality cos2 - - 1-2 is

The correct options are A θ : (8n+1) π/4 B θ : (8n 3) π/4 ...

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

The solution ...

Question

The solution set of inequality $c o s^{2} θ < \frac{1}{2}$ is

{

θ : (8 n + 1) \frac{π}{4} < θ < (8 n + 3) \frac{π}{4}, n ϵ Z

}

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{

θ : (8 n - 3) \frac{π}{4} < θ < (8 n - 1) \frac{π}{4}, n ϵ Z

}

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{

θ : (4 n + 1) \frac{π}{4} < θ < (4 n + 3) \frac{π}{4}, n ϵ Z

}

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None of these

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y = \sqrt{(\frac{1 + c o s 2 θ}{1 - c o s 2 θ})}, \frac{d y}{d x} a t θ = \frac{3 π}{4}

I f y = \sqrt{(\frac{1 + c o s 2 θ}{1 - c o s 2 θ})}, \frac{d y}{d θ} a t θ = \frac{3 π}{4} i s

i f sin θ = - \frac{4}{5}, π < θ < \frac{3 π}{2}, t h e n f i n d

sin 2 θ

cos 2 θ

tan 2 θ

A = {θ : 2 {cos}^{2} + θ + sin θ \leq 2}

B = {θ : π / 2 \leq θ \leq 3 π / 2} .

A \cap B = {θ : π / 2 \leq θ \leq 5 π / 6 o r π \leq θ \leq 3 π / 4}

y = \sqrt{\frac{1 + cos 2 θ}{1 - cos 2 θ}}

\frac{d y}{d θ}

θ = \frac{3 π}{4}

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