If sin2⁡(x+π/4)+3cos⁡2x>0, then

2sin2⁡(x+π/4)+3cos⁡2x>0⇒1−cos⁡(2x+π/2)+3cos⁡2x>0⇒12sin⁡2x+32cos⁡2x>−12⇒cos⁡(2x−π/6)>−12

If sin2⁡(x+π/4)+3cos⁡2x>0, then

2sin2⁡(x+π/4)+3cos⁡2x>0⇒1−cos⁡(2x+π/2)+3cos⁡2x>0⇒12sin⁡2x+32cos⁡2x>−12⇒cos⁡(2x−π/6)>−12

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If $\sin^{2} (x + π / 4) + \sqrt{3} \cos 2 x > 0,$ then

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$\cos (2 x - π / 6) > - 1 / 2$

$\sin (2 x - π / 6) < - 1 / 2$

$\sin (2 x - π / 6) > - 1 / 2$

$\cos (2 x - π / 6) < - 1 / 2$

answer is A.

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Detailed Solution

$\begin{array}{l} 2 \sin^{2} (x + π / 4) + \sqrt{3} \cos 2 x > 0 \\ \Rightarrow 1 - \cos (2 x + π / 2) + \sqrt{3} \cos 2 x > 0 \\ \Rightarrow \frac{1}{2} \sin 2 x + \frac{\sqrt{3}}{2} \cos 2 x > - \frac{1}{2} \\ \Rightarrow \cos (2 x - π / 6) > - \frac{1}{2} \end{array}$