Let tan - 1 ( x ) ( - 2 , 2 ) for x . Then the number of real solutions of the equation 1 + cos ( 2 x ) = 2 ? tan - 1 ( tan x ) in the set ( - 3 2 , - 2 ) ? ( - 2 , 2 ) ( 2 , 3 2 ) is equal to [2023]

Question

Let $\tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})$ for $x \in ℝ$ . Then the number of real solutions of the equation $\sqrt{1 + \cos (2 x)} = \sqrt{2} \tan^{- 1} (\tan x)$ in the

set $(- \frac{3 π}{2}, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{2})$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

$\sqrt{1 + \cos 2 x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow \sqrt{2 \cos^{2} x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow | \cos x | = \tan^{- 1} (\tan x)$

$Number of solutions = Number of intersection points = 3$

Solution

Ans.
(3)

$\sqrt{1 + \cos 2 x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow \sqrt{2 \cos^{2} x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow | \cos x | = \tan^{- 1} (\tan x)$

$Number of solutions = Number of intersection points = 3$

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Solution

Q. Let tan-1(x)∈(-π2,π2) for x∈ℝ. Then the number of real solutions of the equation 1+cos(2x)=2tan-1(tanx) in the set (-3π2,-π2)∪(-π2,π2)∪(π2,3π2) is equal to [2023]

Ans. (3) 1+cos2x=2tan-1(tanx) ⇒ 2cos2x=2tan-1(tanx) ⇒ |cosx|=tan-1(tanx) Number of solutions=Number of intersection points=3

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Ans.
(3)

$\sqrt{1 + \cos 2 x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow \sqrt{2 \cos^{2} x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow | \cos x | = \tan^{- 1} (\tan x)$

$Number of solutions = Number of intersection points = 3$