if 2cos⁡x+6sin⁡x=max(sin⁡θ+cos⁡θ),then x=

max(sin⁡θ+cos⁡θ)=2∴2cos⁡x+6sin⁡x=2Dividing by 22, we getcos⁡x2+32sin⁡x=12⇒cos⁡x−π3=12=cos⁡π3∴x−π3=2nπ±π3⇒x=2nπ+2π3,2nπ.

if 2cos⁡x+6sin⁡x=max(sin⁡θ+cos⁡θ),then x=

max(sin⁡θ+cos⁡θ)=2∴2cos⁡x+6sin⁡x=2Dividing by 22, we getcos⁡x2+32sin⁡x=12⇒cos⁡x−π3=12=cos⁡π3∴x−π3=2nπ±π3⇒x=2nπ+2π3,2nπ.

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if $\sqrt{2} \cos x + \sqrt{6} \sin x = max (\sin θ + \cos θ),$ then $x =$

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$2 n π + \frac{π}{3}$

$2 n π + \frac{2 π}{3}$

$2 n π + \frac{4 π}{3}$

$2 n π$ +1

answer is B.

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Detailed Solution

$max (\sin θ + \cos θ) = \sqrt{2}$

$∴ \sqrt{2} \cos x + \sqrt{6} \sin x = \sqrt{2}$

Dividing by $2 \sqrt{2}$ , we get

$\begin{array}{l} \frac{\cos x}{2} + \frac{\sqrt{3}}{2} \sin x = \frac{1}{2} \Rightarrow \cos (x - \frac{π}{3}) = \frac{1}{2} = \cos \frac{π}{3} \\ ∴ x - \frac{π}{3} = 2 n π \pm \frac{π}{3} \Rightarrow x = 2 n π + \frac{2 π}{3}, 2 n π . \end{array}$