If sec x + cos x = 5/2, where x lies between 0o and 90o, then what is…

We are given,

sec x + cos x = \(\frac{5 }{2 }\)

{ using , sinx = \(\frac{1 }{cosx }\) }

\(\frac{1 }{cosx }\) + cosx = \(\frac{5 }{2 }\)

1 + cos²x = \(\frac{5 cosx }{2 }\) 

2 + 2cos²x - 5cosx = 0

2cos²x - 5cosx + 2 = 0

2cos²x - 4cosx - cosx + 2 = 0

2cosx ( cosx - 2 ) - 1 ( cosx - 2 ) = 0

EIther ( 2cosx - 1 ) = 0  or ( cosx - 2 ) = 0 

cosx - 2 = 0  is not possible

So, ( 2cosx - 1 ) = 0

cosx = \(\frac{1 }{2 }\)

{ using , cos60º = \(\frac{1 }{2 }\) }

So, x = 60º

Now,

sin²x

= sin²60º

= \(\frac{3 }{4 }\)