Find all solutions of the equation in the interval [0, 2𝜋) algebraically. sin2(x) + cos(x) + 1 = 0

Answer 1

sin²(x)+cos(x)+1=0

(1 - cos²x) + cosx +1 = 0

1 - cos²x+ cosx +1 = 0

2 - cos²x+ cosx = 0

0= - 2 + cos²x - cosx

0= cos²x - cosx - 2

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

[If this step confuses, replace "cosx" with a single variable, say "w" for w²-w-2=(w-2)(w+1),

then put cosx back for w.]

This means either cosx=2 or cox=^-1

But since cosx can only be cosx≤1, the only viable solution is cox=^-1

So, x must be π for 0≤x≤2π

Answer 2

sin²x + cosx + 1 = 0

(1 - cos²x) + cosx - 1 = 0

cos²x - cosx = 0

cosx(cosx - 1) = 0

cosx = 0 or cosx = 1

cosx = 0 when x = π/2 or 3π/2

cosx = 1 when x = 0