Sol ve for x:Cos(2sin-1x )=1/9

Sorry.... U r right.......... ........ Cos(2 sin-¹ x) =1/9............................. Put x/1 into the triangle..... Tan-¹ term is tan-¹x/root (x²/1-x²).................................. Cos(cos-¹ [1-(x²/1-x²)/1+(x²/1-x²)])=1/9............. We take property of tan^-1 and substitute Cos inverse in terms of it... Cos(cos-¹ (1-x²-x²/1-x²+x²)]=1/9................ 1-2x²=1/9............ that is xterm cancels in denominator and we cancel cos & cos - ¹ terms... -2x²=1/9-1................................................ -2x²=1-9/9................................................ -2x²=-8/9............................................................... 2x²=8/9............................................... X²=4/9............................................... X=2/3...............................................sorry that I gave u wrong answer... My mistake... ???

OK let me do that again

But the ans ki given 2/3

Hi saumya...i don't know whether u will understand this answer but I'll try to tell u.... Cos(2 sin^-1x)= 1/9 Put sin^-1 x in a triangle and find tan^-1..it will be tan^-1 x/root (x^2 +1)... 2 tan^-1 x/root (x^2 +1)=cos^-1 [(1- (x²/x²+1)) /(1+(x²/x²+1)] When we solv it with respect to the qn... X²+1-x²/x²+1+x²=1/9 X term cancels in numerator. Then it becomes 1/2x²+1=1/9 So 2x²+1=9and 2x²=8 X²=4 X=2