2cos^2x+sinx+1=02cos2x+sinx+1=0 can be written as
2(1-sin^2x)+sinx+1=02(1−sin2x)+sinx+1=0, which can be simplified to
2sin^2x-sinx-3=02sin2x−sinx−3=0
As this is an equation of the type az^2+bz+c=0az2+bz+c=0, where z=sinxz=sinx, a=2a=2, b=-1b=−1 and c=-3c=−3, whose solution is given by (-b+-sqrt(b^2-4ac))/(2a)−b±√b2−4ac2a, we have
sinx=(-(-1)+-sqrt((-1)^2-4*2*(-3)))/(2.2)sinx=−(−1)±√(−1)2−4⋅2⋅(−3)2.2 or
sinx=(1+-sqrt25)/4sinx=1±√254 or
sinx=(1+-5)/4sinx=1±54 i.e. sinx= 3/2 or -1sinx=32or−1
As range of sinxsinx is within [-1,1}, it cannot be 3/232 and -1−1 isonly possibility.
Hence, possible solution within the domain [0, 2pi][0,2π] are
x=(3pi)/2x=3π2
