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

The general solution for -cos(x)=-sin^2(x)-1,(0,2pi) is No Solution for x\in\mathbb{R}

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

- cos (x) = - sin^{2} (x) - 1, (0, 2 π)

No Solution for x \in R

Solution steps

- cos (x) = - sin^{2} (x) - 1, (0, 2 π)

Subtract

- sin^{2} (x) - 1

from both sides

- cos (x) + sin^{2} (x) + 1 = 0

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - 2, (0, 2 π) :

No Solution

cos (x) = 1, (0, 2 π) :

No Solution

Combine all the solutions

No Solution for x \in R

x, y + arctan (\frac{y}{x}) = 0

tan (θ) = - 1, 0 \leq θ \leq 2 π

0.85 = \frac{1 - sin ( x )}{1 + sin ( x )}

sin (2 x) - \frac{2}{2} = 0

y, z = x sin (x y)

What is the general solution for -cos(x)=-sin^2(x)-1,(0,2pi) ?
The general solution for -cos(x)=-sin^2(x)-1,(0,2pi) is No Solution for x\in\mathbb{R}