Calculus Examples

$\cos (2 x) + \sin (x) = 1$ , $[0, 2 π)$

Step 1

Subtract $1$ from both sides of the equation.

$\cos (2 x) + \sin (x) - 1 = 0$

Step 2

Simplify the left side of the equation.

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Step 2.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

$1 - 2 \sin^{2} (x) + \sin (x) - 1 = 0$

Step 2.2

Combine the opposite terms in $1 - 2 \sin^{2} (x) + \sin (x) - 1$ .

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Step 2.2.1

Subtract $1$ from $1$ .

$0 - 2 \sin^{2} (x) + \sin (x) = 0$

Step 2.2.2

Subtract $2 \sin^{2} (x)$ from $0$ .

$- 2 \sin^{2} (x) + \sin (x) = 0$

Step 3

Factor $\sin (x)$ out of $- 2 \sin^{2} (x) + \sin (x)$ .

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Step 3.1

Factor $\sin (x)$ out of $- 2 \sin^{2} (x)$ .

$\sin (x) (- 2 \sin (x)) + \sin (x) = 0$

Step 3.2

Raise $\sin (x)$ to the power of $1$ .

$\sin (x) (- 2 \sin (x)) + \sin (x) = 0$

Step 3.3

Factor $\sin (x)$ out of $\sin^{1} (x)$ .

$\sin (x) (- 2 \sin (x)) + \sin (x) \cdot 1 = 0$

Step 3.4

Factor $\sin (x)$ out of $\sin (x) (- 2 \sin (x)) + \sin (x) \cdot 1$ .

$\sin (x) (- 2 \sin (x) + 1) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sin (x) = 0$

$- 2 \sin (x) + 1 = 0$

Step 5

Set $\sin (x)$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Step 5.2

Solve $\sin (x) = 0$ for $x$ .

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Step 5.2.1

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

$x = \arcsin (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

Step 5.2.3

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - 0$

Step 5.2.4

Subtract $0$ from $π$ .

$x = π$

Step 5.2.5

Find the period of $\sin (x)$ .

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Step 5.2.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 5.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 5.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.6

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 6

Set $- 2 \sin (x) + 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $- 2 \sin (x) + 1$ equal to $0$ .

$- 2 \sin (x) + 1 = 0$

Step 6.2

Solve $- 2 \sin (x) + 1 = 0$ for $x$ .

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$- 2 \sin (x) = - 1$

Step 6.2.2

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ and simplify.

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Step 6.2.2.1

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ .

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $- 2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

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

Step 6.2.2.2.1.2

Divide $\sin (x)$ by $1$ .

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

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 6.2.3

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

$x = \arcsin (\frac{1}{2})$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 6.2.5

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{6}$

Step 6.2.6

Simplify $π - \frac{π}{6}$ .

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Step 6.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 6.2.6.2

Combine fractions.

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Step 6.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 6.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 6.2.6.3

Simplify the numerator.

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Step 6.2.6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 6.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 6.2.7

Find the period of $\sin (x)$ .

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Step 6.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 6.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 6.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 6.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $\sin (x) (- 2 \sin (x) + 1) = 0$ true.

$x = 2 π n, π + 2 π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 8

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

$x = π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 9

Find the values of $n$ that produce a value within the interval $[0, 2 π)$ .

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Step 9.1

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

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Step 9.1.1

Plug in $0$ for $n$ .

$π (0)$

Step 9.1.2

Multiply $π$ by $0$ .

$0$

Step 9.1.3

The interval $[0, 2 π)$ contains $0$ .

$x = 0$

Step 9.2

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

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Step 9.2.1

Plug in $0$ for $n$ .

$\frac{π}{6} + 2 π (0)$

Step 9.2.2

Simplify.

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Step 9.2.2.1

Multiply $2 π (0)$ .

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Step 9.2.2.1.1

Multiply $0$ by $2$ .

$\frac{π}{6} + 0 π$

Step 9.2.2.1.2

Multiply $0$ by $π$ .

$\frac{π}{6} + 0$

Step 9.2.2.2

Add $\frac{π}{6}$ and $0$ .

$\frac{π}{6}$

Step 9.2.3

The interval $[0, 2 π)$ contains $\frac{π}{6}$ .

$x = 0, \frac{π}{6}$

Step 9.3

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

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Step 9.3.1

Plug in $0$ for $n$ .

$\frac{5 π}{6} + 2 π (0)$

Step 9.3.2

Simplify.

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Step 9.3.2.1

Multiply $2 π (0)$ .

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Step 9.3.2.1.1

Multiply $0$ by $2$ .

$\frac{5 π}{6} + 0 π$

Step 9.3.2.1.2

Multiply $0$ by $π$ .

$\frac{5 π}{6} + 0$

Step 9.3.2.2

Add $\frac{5 π}{6}$ and $0$ .

$\frac{5 π}{6}$

Step 9.3.3

The interval $[0, 2 π)$ contains $\frac{5 π}{6}$ .

$x = 0, \frac{π}{6}, \frac{5 π}{6}$

Step 9.4

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

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Step 9.4.1

Plug in $1$ for $n$ .

$π (1)$

Step 9.4.2

Multiply $π$ by $1$ .

$π$

Step 9.4.3

The interval $[0, 2 π)$ contains $π$ .

$x = 0, \frac{π}{6}, \frac{5 π}{6}, π$