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

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

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1+cos^2(x)=0

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

cos (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 = cos (x)

cos (x) = i, cos (x) = - i

cos (x) = i, cos (x) = - i

cos (x) = i :

No Solution

cos (x) = i

No Solution

cos (x) = - i :

No Solution

cos (x) = - i

No Solution

Combine all the solutions

No Solution for x \in R

tan (2 x) = - 1.238

tan (θ) = π

csc^{2} (x) + 2 cot^{2} (x) = 1

3 csc (2 x) + 2 = 0

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

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