What is the general solution for sin^2(x)+cos(x)+1=0,0<= x<= 2pi ?

The general solution for sin^2(x)+cos(x)+1=0,0<= x<= 2pi is x=pi

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sin^2(x)+cos(x)+1=0,0<= x<= 2pi

sin^{2} (x) + cos (x) + 1 = 0, 0 \leq x \leq 2 π

x = π

Solution steps

sin^{2} (x) + cos (x) + 1 = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

2 + cos (x) - cos^{2} (x) = 0

Solve by substitution

cos (x) = - 1, cos (x) = 2

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = 2, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = π

sin (2 x) = \frac{- 1}{3}

cos (2 θ) = 1 + 3 sin (θ)

cos (2 x) - cos (x) = - 1, 0 \leq x < 2 π

sin (x) = - \frac{1}{10}

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π

What is the general solution for sin^2(x)+cos(x)+1=0,0<= x<= 2pi ?
The general solution for sin^2(x)+cos(x)+1=0,0<= x<= 2pi is x=pi