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

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

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cos^2(x)+1=0

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} + 1 = 0

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

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

u^{2} + 1 - 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

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

cos (2 x + \frac{π}{6}) = \frac{1}{2}

(1.54)^{2} = 2 - 2 cos (x)

1 = arctan (x)

4 sec (x) + 8 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

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