Trigonometry Examples

$x + y^{2} = 0$

Step 1

Solve for $y$ .

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Step 1.1

Subtract $x$ from both sides of the equation.

$y^{2} = - x$

Step 1.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- x}$

Step 1.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.3.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{- x}$

Step 1.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{- x}$

Step 1.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{- x}$

$y = - \sqrt{- x}$

$y = \sqrt{- x}$

$y = - \sqrt{- x}$

$y = \sqrt{- x}$

$y = - \sqrt{- x}$

Step 2

Set the radicand in $\sqrt{- x}$ less than $0$ to find where the expression is undefined.

$- x < 0$

Step 3

Divide each term in $- x < 0$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- x < 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{0}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{0}{- 1}$

Step 3.2.2

Divide $x$ by $1$ .

$x > \frac{0}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide $0$ by $- 1$ .

$x > 0$

Step 4

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x > 0$

$(0, \infty)$

Step 5