Since, cos^ x + sin^2 x = 1cosx+sin2x=1
we have: cos^2 x = 1 - sin^2 xcos2x=1−sin2x
So we can replace cos^2 xcos2x in the equation 1 + sinx = 2cos^2x1+sinx=2cos2x by (1- sin^2 x)(1−sin2x)
=>2 (1 - sin^2 x) = sin x +1⇒2(1−sin2x)=sinx+1
or, 2 - 2 sin^2 x = sin x + 12−2sin2x=sinx+1
or, 0=2sin ^2 x + sin x + 1 - 20=2sin2x+sinx+1−2
or, 2sin ^2 x + sin x - 1 = 02sin2x+sinx−1=0
using the quadratic formula:
x = (-b+-sqrt(b^2 - 4ac))/(2a)x=−b±√b2−4ac2a for quadratic equation ax^2+bx+c=0ax2+bx+c=0
we have:
sin x = (-1+-sqrt(1^2 - 4*2*(-1)))/(2*2)sinx=−1±√12−4⋅2⋅(−1)2⋅2
or, sin x = (-1+-sqrt(1 + 8))/4sinx=−1±√1+84
or, sin x = (-1+-sqrt(9))/4sinx=−1±√94
or, sin x = (-1+-3)/4sinx=−1±34
or, sin x = (-1+3)/4, (-1-3)/4sinx=−1+34,−1−34
or, sin x = 1/2, -1sinx=12,−1
Case I:
sin x = 1/2sinx=12
for the condition: 0<=x<=2pi0≤x≤2π
we have:
x= pi/6 or (5pi)/6x=π6or5π6 to get positive value of sinxsinx
Case II:
sin x = -1sinx=−1
we have:
x= (3pi)/2x=3π2 to get negative value of sinxsinx
