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Question

If 2cos3B+3cos4A=3 and 2sin3B3sin4A=0, where 2A and 2B are positive acute angles. If 2A+3B=πJ, then the value of J is

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is B 2
Given,
2cos3B+3cos4A=3
2cos3B=3(1cos4A)
2cos3B=6sin22A (i)

and 2sin3B=3sin4A
2sin3B=6sin2Acos2A (ii)

From (i)/(ii), we get
cos3Bsin3B=sin2Acos2A
cos3Bcos2Asin3Bsin2A=0
cos(2A+3B)=0
2A,2B are positive acute angles
0<2A+3B<5π4
2A+3B=π2J=2

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