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very urgent >>>>>> if 1/6 sinx ,cosx , tanx , are in GP then find the general solution of x
cos^2x=1/6sinx.tanx6cos^3x=sin^2x=1-cos^2x6cos^3x+ cos^2x-1=0put cosx=y then6y^3+y^2-1=0put y=1/26/8 +1/4-1=3/4+1/4-1=1-1=0hence ( 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=1/2cosx=1/2=cospi/3x=pi/3Thanks & RegardsRinkoo GuptaAskIITians Faculty
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