If the sum of all possible values of x in -0- satisfying 2sin-1cos x-3-0 is k-2- then k lies in the interval

Sin(^ 1(cos x))=√(3)2 ⇒^ 1(cos x)=nπ+( 1)^n(π3), n∈ℤ But ^ 1x∈(0,π) ∴^ 1(cos x)=π3,2π3 ⇒cos x =1√(3), 1√(3) ∴ x=α, π α where cosα=1√(3) Required sum =α+π α=π No ...

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Byju's Answer

Standard XII

Mathematics

Inverse Function

If the sum of...

Question

If the sum of all possible values of $x$ in $[0, π]$ satisfying $2 sin ({cot}^{- 1} (cos x)) - \sqrt{3} = 0$ is $\frac{k π}{2},$ then $k$ lies in the interval

[0, 2]

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[0, 4]

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[3, 5]

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[0, 1]

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Solution

The correct option is B $[0, 4]$
$sin ({cot}^{- 1} (cos x)) = \frac{\sqrt{3}}{2}$
$\Rightarrow {cot}^{- 1} (cos x) = n π + (- 1)^{n} (\frac{π}{3}), n \in Z$
But ${cot}^{- 1} x \in (0, π)$
$∴ {cot}^{- 1} (cos x) = \frac{π}{3}, \frac{2 π}{3}$
$\Rightarrow cos x = \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}$
$∴ x = α, π - α$ where $cos α = \frac{1}{\sqrt{3}}$
Required sum $= α + π - α = π$
Now, $π = \frac{k π}{2} \Rightarrow k = 2$

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If the sum of all possible values of x in [0,π] satisfying 2sin(cot−1(cosx))−√3=0 is kπ2, then k lies in the interval

The correct option is B [0,4]sin(cot−1(cos x))=√32 ⇒cot−1(cosx)=nπ+(−1)n(π3), n∈Z But cot−1x∈(0,π) ∴cot−1(cosx)=π3,2π3 ⇒cosx=1√3,−1√3 ∴x=α, π−α where cosα=1√3 Required sum =α+π−α=π Now, π=kπ2⇒k=2

If the sum of all possible values of $x$ in $[0, π]$ satisfying $2 sin ({cot}^{- 1} (cos x)) - \sqrt{3} = 0$ is $\frac{k π}{2},$ then $k$ lies in the interval