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If sinθ6,cosθ and tanθ arein GPthen the general value of θ is 

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a
2±π3,nI
b
2±π6,nI
c
2+(1)nπ3,nI
d
+π3,nI

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detailed solution

Correct option is A

Since,cos2θ=16sinθtanθ

 6cos3θ+cos2θ1=0

As  cos  0  = 1  satisfied the  equation.

 (2cosθ1)3cos2θ+2cosθ+1=0

 cosθ=12 (other values of cosθ are imaginary)  θ=2±π3,nI

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