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

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

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1+sin^2(x)=0

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

1 + u^{2} = 0

1 + u^{2} = 0 : u = i, u = - i

1 + u^{2} = 0

Move

1

to the right side

1 + u^{2} = 0

Subtract

1

from both sides

1 + u^{2} - 1 = 0 - 1

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

Substitute back

u = sin (x)

sin (x) = i, sin (x) = - i

sin (x) = i, sin (x) = - i

sin (x) = i :

No Solution

sin (x) = i

No Solution

sin (x) = - i :

No Solution

sin (x) = - i

No Solution

Combine all the solutions

No Solution for x \in R

2 sin^{2} (x) + 4 cos^{2} (x) = 1

- sin (θ) = \frac{n}{2} - 3

3 sin (θ) + 2 sin^{2} (θ) = - 1, 0 \leq θ < 2 π

3 sin^{2} (x) + 4 sin (x) + 1 = 0

cos (x) = \frac{9}{14}

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