IIT JEE , AIEEE Forum - Mathematics , Trignometry

puneet

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let us use the identity cos²x + sin²x = 1 to solve this question

Now since |cosx| <= 1 and |sinx| <= 1

so, cos³x <= cos²x and sin⁵x <= sin²x

so, 1 = sin⁵x+cos³x <= cos²x + sin²x = 1

so, the equality cos³x = cos²x and sin⁵x = sin²x must hold

Now, cos³x = cos²x implies cosx = 0 or 1

and, sin⁵x = sin²x means sinx = 0 or sinx = 1

Hence the solution is cosx = 0, sinx = 1 or cosx = 1 , sinx = 0

So the value of x is x = (4n+1)

/2 or x = 2n

cheers