A and B are positive acute angles satisfying the equations 3cos2⁡A+2cos2⁡B=4 and 3sin⁡Asin⁡B=2cos⁡B¯cos⁡A, then A+2B is equal to

A and B are positive acute angles satisfying the equations 3cos2A+2cos2B=4 and 3sinAsinB=2cosB¯cosA, then A+2B is equal to

  1. A

    π/4

  2. B

    πβ

  3. C

    π/6

  4. D

    π/2

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    Solution:

    We have,

           3sinAsinB=2cosBcosA 3sinA cosA=2cosBsinBcos2A tanA=sin2B3cos2A

     tanA=sin2Bcos2B×cos2B3cos2A tanA=tan2B×2cos2B13cos2A tanA=tan2B×43cos2A13cos2A tanA=tan2B×33cos2A3cos2A tanA=tan2B×tan2A

     tanAtan2B=1 cosAcos2B=sinAsin2B cos(A+2B)=0A+2B=π2

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