1/6 sin x,cos x,tan x are in G.P., then x=?
(General solution needed)urgent pppls
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1/6 sin x , cos x , tan x are in G.P
cos ^2 x = 1/6/sin x * tan x
cos ^2 x = 1/6 sin x * sin x/cos x
cos ^3 x = 1/6 sin ^2 x
cos ^3 x = 1/6 (1- cos ^2 x)
6 cos ^3 x = 1 -cos ^2 x
6 cos ^3 x +cos ^2 x -1 =0
solve the equation to find x
cos ^2 x = 1/6/sin x * tan x
cos ^2 x = 1/6 sin x * sin x/cos x
cos ^3 x = 1/6 sin ^2 x
cos ^3 x = 1/6 (1- cos ^2 x)
6 cos ^3 x = 1 -cos ^2 x
6 cos ^3 x +cos ^2 x -1 =0
solve the equation to find x
mysticd:
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