What is the general solution for cos^2(x)=-1/3 ?

The general solution for cos^2(x)=-1/3 is No Solution for x\in\mathbb{R}

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cos^2(x)=-1/3

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} = - \frac{1}{3}

u^{2} = - \frac{1}{3} : u = i \frac{1}{3}, u = - i \frac{1}{3}

u^{2} = - \frac{1}{3}

For

x^{2} = f (a)

the solutions are

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

u = - \frac{1}{3}, u = - - \frac{1}{3}

Simplify

- \frac{1}{3} : i \frac{1}{3}

- \frac{1}{3}

Apply radical rule:

- a = - 1 a

- \frac{1}{3} = - 1 \frac{1}{3}

= - 1 \frac{1}{3}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{3}

Simplify

- - \frac{1}{3} : - i \frac{1}{3}

- - \frac{1}{3}

Simplify

- \frac{1}{3} : i \frac{1}{3}

- \frac{1}{3}

Apply radical rule:

- a = - 1 a

- \frac{1}{3} = - 1 \frac{1}{3}

= - 1 \frac{1}{3}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{3}

= - i \frac{1}{3}

u = i \frac{1}{3}, u = - i \frac{1}{3}

Substitute back

u = cos (x)

cos (x) = i \frac{1}{3}, cos (x) = - i \frac{1}{3}

cos (x) = i \frac{1}{3}, cos (x) = - i \frac{1}{3}

cos (x) = i \frac{1}{3} :

No Solution

cos (x) = i \frac{1}{3}

No Solution

cos (x) = - i \frac{1}{3} :

No Solution

cos (x) = - i \frac{1}{3}

No Solution

Combine all the solutions

No Solution for x \in R

What is the general solution for cos^2(x)=-1/3 ?
The general solution for cos^2(x)=-1/3 is No Solution for x\in\mathbb{R}