Question

Let 0 < A,B < $\frac{\pi }{2}$ satisfying the equation $3si{n}^{2}A+2si{n}^{2}B=1$ and 3sin2A - 2sin2B = 0 then A + 2B is equal to

• A. π
• B. π/2
• C. π/4
• D. 2π

Answer 1

The correct option is B

π/2

$3si{n}^{2}A=1-2si{n}^{2}B=cos2B$

Now cos(A+ 2B) = cosA cos2B - sinA sin2B

= 3cosA.$si{n}^{2}A-\frac{3}{2}.sinAsin2A$

= 3cosA.$si{n}^{2}A-3si{n}^{2}A.cosA=0$

$\Rightarrow$ A + 2B = $\frac{\pi }{2}or\frac{3\pi }{2}$

Given that 0 < A < $\frac{\pi }{2}$ and 0 < B < $\frac{\pi }{2}$

$\Rightarrow$ 0 < A + 2B < $\pi +\frac{\pi }{2}$ Hence A + 2B = $\frac{\pi }{2}$