IIT JEE , AIEEE Forum - Mathematics , Trignometry

sprinkle

Vahiniprasad!

Suppose cos x = y

=> y

[-1, 1]

now 2y, |y| and 1-3y^2 are in G.P.

=> |y|^2 = 2y(1-3y^2)
=> y^2 = 2y - 6y^3
=> 6y^3 + y^2 - 2y = 0
solving we get, y = 0, 1/2, -2/3

=> cos x = 0, 1/2, -2/3 (though cos x can't be 0 as then 0, 0, 1 are not in G.P.)
=> x = (2k pi + pi/2), (2k pi - pi/2), (2k pi + pi/3), (2k pi - pi/3),
(2k pi + (sqrt(3) - 1) pi), (2k pi - (sqrt(3) - 1) pi) where k is an integer

so answer is none of ur values given. rather it can be infinity.

Please buzz me if any doubts!

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