Solution(s) of the equation sin⁡x2+cos⁡x2=2sin⁡x is/are

sin⁡x2+cos⁡x2=2sin⁡x⇒1+2sin⁡x2cos⁡x2=2sin2⁡x⇒1+sin⁡x=2sin2⁡x⇒(2sin⁡x+1)(sin⁡x−1)=0⇒sin⁡x=−12 or sin⁡x=1⇒x=nπ+(−1)n−π6 or x=nπ+(−1)nπ2,n∈Z⇒x=nπ+(−1)n+1π62 or x=nπ+(−1)nπ22,n∈Z

Solution(s) of the equation sin⁡x2+cos⁡x2=2sin⁡x is/are

sin⁡x2+cos⁡x2=2sin⁡x⇒1+2sin⁡x2cos⁡x2=2sin2⁡x⇒1+sin⁡x=2sin2⁡x⇒(2sin⁡x+1)(sin⁡x−1)=0⇒sin⁡x=−12 or sin⁡x=1⇒x=nπ+(−1)n−π6 or x=nπ+(−1)nπ2,n∈Z⇒x=nπ+(−1)n+1π62 or x=nπ+(−1)nπ22,n∈Z

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Solution(s) of the equation $\sin (\frac{\sqrt{x}}{2}) + \cos (\frac{\sqrt{x}}{2}) = \sqrt{2} \sin \sqrt{x} is/are$

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$x = {(n π + (- 1)^{n + 1} \frac{π}{3})}^{2}, n \in Z$

$x = {(2 n π + (- 1)^{n} \frac{π}{2})}^{2}, n \in Z$

$x = {(n π + (- 1)^{n + 1} \frac{π}{6})}^{2}, n \in Z$

$x = {(n π + (- 1)^{n + 1} \frac{π}{2})}^{2}, n \in Z$

answer is C.

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Detailed Solution

$\sin (\frac{\sqrt{x}}{2}) + \cos (\frac{\sqrt{x}}{2}) = \sqrt{2} \sin \sqrt{x}$

$\begin{array}{l} \Rightarrow 1 + 2 \sin \frac{\sqrt{x}}{2} \cos \frac{\sqrt{x}}{2} = 2 \sin^{2} \sqrt{x} \\ \Rightarrow 1 + \sin \sqrt{x} = 2 \sin^{2} \sqrt{x} \\ \Rightarrow (2 \sin \sqrt{x} + 1) (\sin \sqrt{x} - 1) = 0 \\ \Rightarrow \sin \sqrt{x} = - \frac{1}{2} or \sin \sqrt{x} = 1 \end{array}$

$\begin{array}{l} \Rightarrow \sqrt{x} = n π + (- 1)^{n} (\frac{- π}{6}) or \sqrt{x} = n π + (- 1)^{n} \frac{π}{2}, n \in Z \\ \Rightarrow x = {(n π + (- 1)^{n + 1} \frac{π}{6})}^{2} or x = {(n π + (- 1)^{n} \frac{π}{2})}^{2}, n \in Z \end{array}$