If $2 cos 3 B + 3 cos 4 A = 3$ and $2 sin 3 B - 3 sin 4 A = 0$ , where $2 A$ and $2 B$ are positive acute angles. If $2 A + 3 B = \frac{π}{J}$ , then the value of $J$ is

1

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2

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Solution

The correct option is B $2$
Given,
$2 cos 3 B + 3 cos 4 A = 3$
$\Rightarrow 2 cos 3 B = 3 (1 - cos 4 A)$
$\Rightarrow 2 cos 3 B = 6 {sin}^{2} 2 A \dots (i)$

and $2 sin 3 B = 3 sin 4 A$
$\Rightarrow 2 sin 3 B = 6 sin 2 A cos 2 A \dots (i i)$

From $(i) / (i i)$ , we get
$\Rightarrow \frac{cos 3 B}{sin 3 B} = \frac{sin 2 A}{cos 2 A}$
$\Rightarrow cos 3 B cos 2 A - sin 3 B sin 2 A = 0$
$\Rightarrow cos (2 A + 3 B) = 0$
$∵ 2 A, 2 B$ are positive acute angles
$∴ 0 < 2 A + 3 B < \frac{5 π}{4}$
$\Rightarrow 2 A + 3 B = \frac{π}{2} \Rightarrow J = 2$

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