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Q.
If A and B are positive acute angles satisfying the equations $3 \, \sin^2 \, A + 2 \, \sin^2 \, B = 1$ and $3 \, \sin \, 2A - 2 \, \sin \, 2B = 0$, then A + 2B =
Trigonometric Functions
Solution:
Given $3 \sin^{2} A + 2 \sin^{2} B = 1$
$ \Rightarrow 3 \sin^{2} A = 1 - 2 \sin^{2} B $
$\Rightarrow 3 \sin^{2} A = \cos2 B $
$ \therefore \cos\left(A + 2 B\right) = \cos A \cos2B - \sin A \sin2B $
$= \left(\cos A\right) \left(3 \sin^{2} A\right) - \sin A \left(\frac{3}{2} \sin 2A\right)$
$ \left(\because3 \sin2A - 2 \sin2 B = 0 \therefore \sin 2B = \frac{2}{3} \sin2A\right) $
$= 3\sin^{2} A \cos A - \frac{3}{2} \left(\sin A\right) \left(2\sin A \cos A\right) = 0$
$ \Rightarrow A + 2B = \frac{\pi}{2} . $