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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 2A−2 sin 2B=0, then A+2B=
$$\mathrm{If}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{acute}\:\mathrm{positive}\:\mathrm{angles} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{equations}\: \\ $$$$\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} {A}+\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} {B}=\mathrm{1}\:\mathrm{and}\: \\ $$$$\mathrm{3}\:\mathrm{sin}\:\mathrm{2}{A}−\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{B}=\mathrm{0},\:\mathrm{then}\:{A}+\mathrm{2}{B}= \\ $$
Commented by john santu last updated on 25/Jan/20
(∗) 3((1/2)−(1/2)cos 2A)+2((1/2)−(1/2)cos 2B)=1  (5/2)−(3/2)cos 2A−cos 2B=1  3cos 2A+2cos 2B=3  9cos^2 2A+12cos 2Acos 2B+4cos^2 2B=9  (∗∗) 3sin 2A −2sin 2B =0  9sin^2 2A−12sin 2Asin 2B+4sin^2 2B=0  (∗)+(∗∗)  13+12cos (2A+2B)=9  cos (2A+2B)=−(1/3)
$$\left(\ast\right)\:\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{A}\right)+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{B}\right)=\mathrm{1} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{A}−\mathrm{cos}\:\mathrm{2}{B}=\mathrm{1} \\ $$$$\mathrm{3cos}\:\mathrm{2}{A}+\mathrm{2cos}\:\mathrm{2}{B}=\mathrm{3} \\ $$$$\mathrm{9cos}\:^{\mathrm{2}} \mathrm{2}{A}+\mathrm{12cos}\:\mathrm{2}{A}\mathrm{cos}\:\mathrm{2}{B}+\mathrm{4cos}\:^{\mathrm{2}} \mathrm{2}{B}=\mathrm{9} \\ $$$$\left(\ast\ast\right)\:\mathrm{3sin}\:\mathrm{2}{A}\:−\mathrm{2sin}\:\mathrm{2}{B}\:=\mathrm{0} \\ $$$$\mathrm{9sin}\:^{\mathrm{2}} \mathrm{2}{A}−\mathrm{12sin}\:\mathrm{2}{A}\mathrm{sin}\:\mathrm{2}{B}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{2}{B}=\mathrm{0} \\ $$$$\left(\ast\right)+\left(\ast\ast\right) \\ $$$$\mathrm{13}+\mathrm{12cos}\:\left(\mathrm{2}{A}+\mathrm{2}{B}\right)=\mathrm{9} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}{A}+\mathrm{2}{B}\right)=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by john santu last updated on 25/Jan/20
A+B = (1/2)cos^(−1) (−(1/3))≈54.7^o
$${A}+{B}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right)\approx\mathrm{54}.\mathrm{7}^{{o}} \\ $$

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