High school mathematics, trigonometric equation Finding the solution set of the equation 2Sin & sup2; X + cosx-1 = 0

High school mathematics, trigonometric equation Finding the solution set of the equation 2Sin & sup2; X + cosx-1 = 0

2sin²x+cosx-1=0
2(1-cos^2x)+cosx-1=0
2cos^2x-cosx-1=0
(2cosx+1)(cosx-1)=0
So, cosx = - 1 / 2 or cosx = 1
That is, x = 2kpi + 2pai / 3 or 2kpi + 4pai / 3 or 2kpi

High one mathematics, trigonometric equation
4cos²x-2sinxcosx-1=0

4cos²x-2sinxcosx-(sin²x+cos²x)=0
3cos²x-2sinxcosx-sin²x=0
(cosx-sinx)(3cosx+sinx)=0
sinx-cosx=0 3cosx+sinx=0
tanx=1 tanx=-3
x=kπ+π/2 x=kπ-arctan3 k∈Z

The solution set of the equation SiNx + cosx = radical 2 / 2 x belonging to [- π, π] is___

（sinx+cosx）^2=1/2
(sinx)^2+(cosx)^2+2sinx*cosx=1/2
Because (SiNx) ^ 2 + (cosx) ^ 2 = 1,2sinx * cosx = sin2x
So sin2x = - 1 / 2
So 2x = 4 / 3 π + 2K π or 5 / 3 π + 2K π
So x = 2 / 3 π + K π or 5 / 6 π + K π
And because x belongs to [- π, π]
So x = 2 / 3 π or 5 / 6 π or - 1 / 3 π or - 1 / 6 π