Consider the following statements/equations
$(i) 2 \sin (3 x + \frac{π}{4}) = \sqrt{1 + 8 \sin 2 {xcos}^{2} 2 x} - - - - (P)$
$(ii) \tan (y + 20^{\circ}) = \tan (y - 10^{\circ}) \tan y \cdot \tan (y + 10^{\circ}) - - - - - (Q)$
(iii)Let $β$ be number of points of intersections of the curves
$y = \cos x and y = \sin 3 x for - \frac{π}{2} \leq x \leq \frac{π}{2} - - - - (R)$

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General solution of (P) $x = \frac{nπ}{2} + (- 1)^{n} \frac{π}{12}, (n \in z)$

If $α$ is the smallest positive angle satisfying (Q) is $\frac{π}{3}$

$\tan^{2} α + 2 \sin 2 x + β = 6$ (where x is solution of (P))

Points of intersection of (R) are $(\frac{π}{4}, \frac{1}{\sqrt{2}}), (\frac{π}{8}, \cos \frac{π}{8}), (- \frac{3 π}{8}, \cos \frac{3 π}{8})$

answer is A, B, D.

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Detailed Solution

$(A) \sqrt{2} (\sin 3 x + \cos 3 x) = \sqrt{1 + 2 (\sin 6 x + \sin 2 x)}$
$s.b.s \Rightarrow \sin 2 x = \frac{1}{2}, x = \frac{nπ}{2} + (- 1)^{n} \frac{π}{12}$
$(B) \frac{\sin (y + 20^{\circ}) \cdot \cos y}{\cos (y + 20^{\circ}) \cdot \sin y} = \frac{\sin (y - 10^{\circ}) \cdot \sin (y + 10^{\circ})}{\cos (y - 10^{\circ}) \cdot \cos (y + 10^{\circ})} u s e c o m p o d e n d o & d i v i d e n d o, r u l e \Leftrightarrow \frac{\sin (2 y + 20 °)}{\sin 20 °} = \frac{\cos 20 °}{- \cos 2 y} \Rightarrow 2 \sin (2 y + 20 °) \cos 2 y = - 2 \sin 20 ° \cos 20 ° \sin (4 y + 20 °) + \sin 20 ° = - \sin 40 ° \sin (4 y + 20 °) = - (\sin 40 ° + \sin 20 °) \sin (4 y + 20 °) = - 2 \sin 30 ° \cos 10 ° = - \cos 10 ° \sin (4 y + 20 °) = \sin 260 ° \Rightarrow 4 y = 240 ∴ y = 60$
$(C) 3 + 2 \cdot \frac{1}{2} + 3 = 7$
(D) Points of Intersection of (C) are $(\frac{π}{4}, \frac{1}{\sqrt{2}}), (\frac{π}{8}, \cos \frac{π}{8}), (- \frac{3 π}{8}, \cos \frac{3 π}{8})$
$((\sin 3 x = \cos x) \Rightarrow 3 x = nπ + (- 1)^{n} (\frac{π}{2} - x) for n = 0 \Rightarrow x = \frac{π}{8}, for n = 1 \Rightarrow x = \frac{π}{4}, for n = - 2, x = - \frac{3 π}{8})$