The range of all values of 'θ'such that sin3θ−cos3θ>0

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The range of all values of $' θ'$ such that $\sin 3 θ - \cos 3 θ > 0$

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$2 n π + (3 π / 4) \leq θ \leq 2 n π + (5 π / 4)$

$2 n π + \frac{π}{12} < θ < 2 n π + \frac{5 π}{12}$

$\frac{n π}{3} - \frac{π}{6} < θ < \frac{n π}{3} + \frac{π}{6}$

$2 n π - \frac{π}{6} < θ < 2 n π + \frac{π}{6}$

answer is B.

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Detailed Solution

$W e h a v e \sin 3 θ - \cos 3 θ > 0 \Rightarrow \frac{1}{\sqrt{2}} \sin 3 θ - \frac{1}{\sqrt{2}} \cos 3 θ > 0 \Rightarrow \sin (3 θ - \frac{π}{4}) > 0 \Rightarrow 0 < 3 θ - \frac{π}{4} < π \Rightarrow \frac{π}{4} < 3 θ < \frac{5 π}{4} \Rightarrow \frac{π}{12} < θ < \frac{5 π}{12} \Rightarrow 2 nπ + \frac{π}{12} < θ < 2 nπ + \frac{5 π}{12}, n \in I$