The number of solutions of the equation cos x=sin^2x,x belongs to 0,6 pi

Question

Aditya Gupta · Answer

dear student, first write the given eqn as C^2+C – 1=0 where C= cosx. now note that the period of the function f(x)= C^2+C – 1 is 2pi. Hence, we only need to find the no of solns in (0, 2pi), and simply triple that to find the no of solns in (0, 6pi). Now on setting C^2+C – 1=0, we get C= (– 1 ± sqrt5)/2. as (– 1 – sqrt5)/2 is less than – 1, we reject it coz C lies in [-1,1]. so we get only C= (– 1+sqrt5)/2. Now, if you draw the graph of y= cosx and y= (– 1+sqrt5)/2, they clearly intersect at only 2 points in [0, 2pi]. Hence, total no of solns in [0, 6pi] is 3*2 = 6 KINDLY APPROVE :))

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The number of solutions of the equation cos x=sin^2x,x belongs to 0,6 pi

The number of solutions of the equation cos x=sin^2x,x belongs to 0,6 pi

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