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

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

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sin^2(x)=-1/2

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

sin (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 = sin (x)

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

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

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

No Solution

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

No Solution

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

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

4 tan^{2} (θ) = 11 tan (θ) + 3

6 tan^{2} (x) - 2 cos^{2} (x) = cos (2 x)

2 cos^{2} (x) - 3 sin (x) - 3 = 0

6 tan (θ) + 5 = 0

x, y = arcsin (x)

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