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

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

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cos^2(x)=-1/2

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

For

x^{2} = f (a)

the solutions are

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

u = - \frac{1}{2}, u = - - \frac{1}{2}

Simplify

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

- \frac{1}{2} = - 1 \frac{1}{2}

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

Simplify

- - \frac{1}{2} : - i \frac{1}{2}

- - \frac{1}{2}

Simplify

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

- \frac{1}{2} = - 1 \frac{1}{2}

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

= - i \frac{1}{2}

u = i \frac{1}{2}, u = - i \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = i \frac{1}{2} :

No Solution

cos (x) = i \frac{1}{2}

No Solution

cos (x) = - i \frac{1}{2} :

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

sin (6 x) = - \frac{1}{2}

2 sin^{2} (θ) + 1 = 3 sin (θ)

sin^{2} (x) = \frac{1}{2}, 0 \leq x \leq 2 π

sin (3 θ) = 1, 0 \leq θ < 2 π

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

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