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}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

α, sin^{2} (3 α) = 0.56

(sin (x) - cos (x))^{2} = 1^{2}

tan (\frac{π}{2} - x) + 2 tan (\frac{π}{2} + x) = 1

\frac{11.0994}{cos ( x )} = \frac{3.5355}{sin ( x )}

(\frac{sec ( x )}{cos ( x )})^{2} = tan^{2} (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}