Calculus Examples

${(\sin (π x) + \cos (π y))}^{2} = 2$

Step 1

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

$\sin (π x) + \cos (π y) = \pm \sqrt{2}$

Step 2

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

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Step 2.1

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

$\sin (π x) + \cos (π y) = \sqrt{2}$

Step 2.2

Subtract $\cos (π y)$ from both sides of the equation.

$\sin (π x) = \sqrt{2} - \cos (π y)$

Step 2.3

Rewrite the equation as $\sqrt{2} - \cos (π y) = \sin (π x)$ .

$\sqrt{2} - \cos (π y) = \sin (π x)$

Step 2.4

Subtract $\sqrt{2}$ from both sides of the equation.

$- \cos (π y) = \sin (π x) - \sqrt{2}$

Step 2.5

Divide each term in $- \cos (π y) = \sin (π x) - \sqrt{2}$ by $- 1$ and simplify.

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Step 2.5.1

Divide each term in $- \cos (π y) = \sin (π x) - \sqrt{2}$ by $- 1$ .

$\frac{- \cos (π y)}{- 1} = \frac{\sin (π x)}{- 1} + \frac{- \sqrt{2}}{- 1}$

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Dividing two negative values results in a positive value.

$\frac{\cos (π y)}{1} = \frac{\sin (π x)}{- 1} + \frac{- \sqrt{2}}{- 1}$

Step 2.5.2.2

Divide $\cos (π y)$ by $1$ .

$\cos (π y) = \frac{\sin (π x)}{- 1} + \frac{- \sqrt{2}}{- 1}$

Step 2.5.3

Simplify the right side.

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Step 2.5.3.1

Simplify each term.

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Step 2.5.3.1.1

Move the negative one from the denominator of $\frac{\sin (π x)}{- 1}$ .

$\cos (π y) = - 1 \cdot \sin (π x) + \frac{- \sqrt{2}}{- 1}$

Step 2.5.3.1.2

Rewrite $- 1 \cdot \sin (π x)$ as $- \sin (π x)$ .

$\cos (π y) = - \sin (π x) + \frac{- \sqrt{2}}{- 1}$

Step 2.5.3.1.3

Dividing two negative values results in a positive value.

$\cos (π y) = - \sin (π x) + \frac{\sqrt{2}}{1}$

Step 2.5.3.1.4

Divide $\sqrt{2}$ by $1$ .

$\cos (π y) = - \sin (π x) + \sqrt{2}$

Step 2.6

Take the inverse cosine of both sides of the equation to extract $x$ from inside the cosine.

$π y = \arccos (- \sin (π x) + \sqrt{2})$

Step 2.7

Rewrite the equation as $\arccos (- \sin (π x) + \sqrt{2}) = π y$ .

$\arccos (- \sin (π x) + \sqrt{2}) = π y$

Step 2.8

Take the inverse arccosine of both sides of the equation to extract $\sin (π x)$ from inside the arccosine.

$- \sin (π x) + \sqrt{2} = \cos (π y)$

Step 2.9

Subtract $\sqrt{2}$ from both sides of the equation.

$- \sin (π x) = \cos (π y) - \sqrt{2}$

Step 2.10

Divide each term in $- \sin (π x) = \cos (π y) - \sqrt{2}$ by $- 1$ and simplify.

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Step 2.10.1

Divide each term in $- \sin (π x) = \cos (π y) - \sqrt{2}$ by $- 1$ .

$\frac{- \sin (π x)}{- 1} = \frac{\cos (π y)}{- 1} + \frac{- \sqrt{2}}{- 1}$

Step 2.10.2

Simplify the left side.

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Step 2.10.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (π x)}{1} = \frac{\cos (π y)}{- 1} + \frac{- \sqrt{2}}{- 1}$

Step 2.10.2.2

Divide $\sin (π x)$ by $1$ .

$\sin (π x) = \frac{\cos (π y)}{- 1} + \frac{- \sqrt{2}}{- 1}$

Step 2.10.3

Simplify the right side.

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Step 2.10.3.1

Simplify each term.

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Step 2.10.3.1.1

Move the negative one from the denominator of $\frac{\cos (π y)}{- 1}$ .

$\sin (π x) = - 1 \cdot \cos (π y) + \frac{- \sqrt{2}}{- 1}$

Step 2.10.3.1.2

Rewrite $- 1 \cdot \cos (π y)$ as $- \cos (π y)$ .

$\sin (π x) = - \cos (π y) + \frac{- \sqrt{2}}{- 1}$

Step 2.10.3.1.3

Dividing two negative values results in a positive value.

$\sin (π x) = - \cos (π y) + \frac{\sqrt{2}}{1}$

Step 2.10.3.1.4

Divide $\sqrt{2}$ by $1$ .

$\sin (π x) = - \cos (π y) + \sqrt{2}$

Step 2.11

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$π x = \arcsin (- \cos (π y) + \sqrt{2})$

Step 2.12

Divide each term in $π x = \arcsin (- \cos (π y) + \sqrt{2})$ by $π$ and simplify.

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Step 2.12.1

Divide each term in $π x = \arcsin (- \cos (π y) + \sqrt{2})$ by $π$ .

$\frac{π x}{π} = \frac{\arcsin (- \cos (π y) + \sqrt{2})}{π}$

Step 2.12.2

Simplify the left side.

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Step 2.12.2.1

Cancel the common factor of $π$ .

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Step 2.12.2.1.1

Cancel the common factor.

$\frac{π x}{π} = \frac{\arcsin (- \cos (π y) + \sqrt{2})}{π}$

Step 2.12.2.1.2

Divide $x$ by $1$ .

$x = \frac{\arcsin (- \cos (π y) + \sqrt{2})}{π}$

Step 2.13

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

$\sin (π x) + \cos (π y) = - \sqrt{2}$

Step 2.14

Subtract $\cos (π y)$ from both sides of the equation.

$\sin (π x) = - \sqrt{2} - \cos (π y)$

Step 2.15

Rewrite the equation as $- \sqrt{2} - \cos (π y) = \sin (π x)$ .

$- \sqrt{2} - \cos (π y) = \sin (π x)$

Step 2.16

Add $\sqrt{2}$ to both sides of the equation.

$- \cos (π y) = \sin (π x) + \sqrt{2}$

Step 2.17

Divide each term in $- \cos (π y) = \sin (π x) + \sqrt{2}$ by $- 1$ and simplify.

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Step 2.17.1

Divide each term in $- \cos (π y) = \sin (π x) + \sqrt{2}$ by $- 1$ .

$\frac{- \cos (π y)}{- 1} = \frac{\sin (π x)}{- 1} + \frac{\sqrt{2}}{- 1}$

Step 2.17.2

Simplify the left side.

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Step 2.17.2.1

Dividing two negative values results in a positive value.

$\frac{\cos (π y)}{1} = \frac{\sin (π x)}{- 1} + \frac{\sqrt{2}}{- 1}$

Step 2.17.2.2

Divide $\cos (π y)$ by $1$ .

$\cos (π y) = \frac{\sin (π x)}{- 1} + \frac{\sqrt{2}}{- 1}$

Step 2.17.3

Simplify the right side.

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Step 2.17.3.1

Simplify each term.

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Step 2.17.3.1.1

Move the negative one from the denominator of $\frac{\sin (π x)}{- 1}$ .

$\cos (π y) = - 1 \cdot \sin (π x) + \frac{\sqrt{2}}{- 1}$

Step 2.17.3.1.2

Rewrite $- 1 \cdot \sin (π x)$ as $- \sin (π x)$ .

$\cos (π y) = - \sin (π x) + \frac{\sqrt{2}}{- 1}$

Step 2.17.3.1.3

Move the negative one from the denominator of $\frac{\sqrt{2}}{- 1}$ .

$\cos (π y) = - \sin (π x) - 1 \cdot \sqrt{2}$

Step 2.17.3.1.4

Rewrite $- 1 \cdot \sqrt{2}$ as $- \sqrt{2}$ .

$\cos (π y) = - \sin (π x) - \sqrt{2}$

Step 2.18

Take the inverse cosine of both sides of the equation to extract $x$ from inside the cosine.

$π y = \arccos (- \sin (π x) - \sqrt{2})$

Step 2.19

Rewrite the equation as $\arccos (- \sin (π x) - \sqrt{2}) = π y$ .

$\arccos (- \sin (π x) - \sqrt{2}) = π y$

Step 2.20

Take the inverse arccosine of both sides of the equation to extract $\sin (π x)$ from inside the arccosine.

$- \sin (π x) - \sqrt{2} = \cos (π y)$

Step 2.21

Add $\sqrt{2}$ to both sides of the equation.

$- \sin (π x) = \cos (π y) + \sqrt{2}$

Step 2.22

Divide each term in $- \sin (π x) = \cos (π y) + \sqrt{2}$ by $- 1$ and simplify.

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Step 2.22.1

Divide each term in $- \sin (π x) = \cos (π y) + \sqrt{2}$ by $- 1$ .

$\frac{- \sin (π x)}{- 1} = \frac{\cos (π y)}{- 1} + \frac{\sqrt{2}}{- 1}$

Step 2.22.2

Simplify the left side.

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Step 2.22.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (π x)}{1} = \frac{\cos (π y)}{- 1} + \frac{\sqrt{2}}{- 1}$

Step 2.22.2.2

Divide $\sin (π x)$ by $1$ .

$\sin (π x) = \frac{\cos (π y)}{- 1} + \frac{\sqrt{2}}{- 1}$

Step 2.22.3

Simplify the right side.

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Step 2.22.3.1

Simplify each term.

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Step 2.22.3.1.1

Move the negative one from the denominator of $\frac{\cos (π y)}{- 1}$ .

$\sin (π x) = - 1 \cdot \cos (π y) + \frac{\sqrt{2}}{- 1}$

Step 2.22.3.1.2

Rewrite $- 1 \cdot \cos (π y)$ as $- \cos (π y)$ .

$\sin (π x) = - \cos (π y) + \frac{\sqrt{2}}{- 1}$

Step 2.22.3.1.3

Move the negative one from the denominator of $\frac{\sqrt{2}}{- 1}$ .

$\sin (π x) = - \cos (π y) - 1 \cdot \sqrt{2}$

Step 2.22.3.1.4

Rewrite $- 1 \cdot \sqrt{2}$ as $- \sqrt{2}$ .

$\sin (π x) = - \cos (π y) - \sqrt{2}$

Step 2.23

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$π x = \arcsin (- \cos (π y) - \sqrt{2})$

Step 2.24

Divide each term in $π x = \arcsin (- \cos (π y) - \sqrt{2})$ by $π$ and simplify.

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Step 2.24.1

Divide each term in $π x = \arcsin (- \cos (π y) - \sqrt{2})$ by $π$ .

$\frac{π x}{π} = \frac{\arcsin (- \cos (π y) - \sqrt{2})}{π}$

Step 2.24.2

Simplify the left side.

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Step 2.24.2.1

Cancel the common factor of $π$ .

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Step 2.24.2.1.1

Cancel the common factor.

$\frac{π x}{π} = \frac{\arcsin (- \cos (π y) - \sqrt{2})}{π}$

Step 2.24.2.1.2

Divide $x$ by $1$ .

$x = \frac{\arcsin (- \cos (π y) - \sqrt{2})}{π}$

Step 2.25

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

$x = \frac{\arcsin (- \cos (π y) + \sqrt{2})}{π}$

$x = \frac{\arcsin (- \cos (π y) - \sqrt{2})}{π}$

$x = \frac{\arcsin (- \cos (π y) + \sqrt{2})}{π}$

$x = \frac{\arcsin (- \cos (π y) - \sqrt{2})}{π}$