If A and B are acute positive angles satisfying the equations 3 sin^2 A+2 sin^2 B=1 and 3 sin 2A−2 sin 2B=0, then A+2B=


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Question Number 79456 by Vishal Sharma last updated on 25/Jan/20

$If A and B are acute positive angles$ $satisfying the equations$ $3 \sin^{2} A + 2 \sin^{2} B = 1 and$ $3 \sin 2 A - 2 \sin 2 B = 0, then A + 2 B =$
Commented by john santu last updated on 25/Jan/20

$() 3 (\frac{1}{2} - \frac{1}{2} \cos 2 A) + 2 (\frac{1}{2} - \frac{1}{2} \cos 2 B) = 1$ $\frac{5}{2} - \frac{3}{2} \cos 2 A - \cos 2 B = 1$ $3 \cos 2 A + 2 \cos 2 B = 3$ $9 \cos^{2} 2 A + 12 \cos 2 A \cos 2 B + 4 \cos^{2} 2 B = 9$ $( ) 3 \sin 2 A - 2 \sin 2 B = 0$ $9 \sin^{2} 2 A - 12 \sin 2 A \sin 2 B + 4 \sin^{2} 2 B = 0$ $() + (* *)$ $13 + 12 \cos (2 A + 2 B) = 9$ $\cos (2 A + 2 B) = - \frac{1}{3}$
Commented by john santu last updated on 25/Jan/20

$A + B = \frac{1}{2} \cos^{- 1} (- \frac{1}{3}) \approx 54 . 7^{o}$
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