Rinkoo Gupta
Last Activity: 11 Years ago
Since 1/6 sinx, cosx and tanx are in G.P. So
cos^2x=1/6sinx.tanx6cos^3x=sin^2x=1-cos^2x
6cos^3x+ cos^2x-1=0
put cosx=y then
6y^3+y^2-1=0
put y=1/2
6/8 +1/4-1=3/4+1/4-1=1-1=0
hence ( y-1/2) is the factor of above eq.
hence the factors are
(y-1/2)(6y^2+4y+2)=0
=>(y-1/2)=0 and second factor have imaginary roots.
so y=1/2
cosx=1/2=cospi/3
x=pi/3
Thanks & Regards
Rinkoo Gupta
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