Since, #cos^ x + sin^2 x = 1#
we have: #cos^2 x = 1 - sin^2 x#
So we can replace #cos^2 x# in the equation #1 + sinx = 2cos^2x# by #(1- sin^2 x)#
#=>2 (1 - sin^2 x) = sin x +1#
or, #2 - 2 sin^2 x = sin x + 1#
or, #0=2sin ^2 x + sin x + 1 - 2#
or, #2sin ^2 x + sin x - 1 = 0#
using the quadratic formula:
#x = (-b+-sqrt(b^2 - 4ac))/(2a)# for quadratic equation #ax^2+bx+c=0#
we have:
#sin x = (-1+-sqrt(1^2 - 4*2*(-1)))/(2*2)#
or, #sin x = (-1+-sqrt(1 + 8))/4#
or, #sin x = (-1+-sqrt(9))/4#
or, #sin x = (-1+-3)/4#
or, #sin x = (-1+3)/4, (-1-3)/4#
or, #sin x = 1/2, -1#
Case I:
#sin x = 1/2#
for the condition: #0<=x<=2pi#
we have:
#x= pi/6 or (5pi)/6# to get positive value of #sinx#
Case II:
#sin x = -1#
we have:
#x= (3pi)/2# to get negative value of #sinx#