Precalculus Examples

$\sin (2 x) = - 13 \cos (x)$

Step 1

Add $13 \cos (x)$ to both sides of the equation.

$\sin (2 x) + 13 \cos (x) = 0$

Step 2

Apply the sine double-angle identity.

$2 \sin (x) \cos (x) + 13 \cos (x) = 0$

Step 3

Factor $\cos (x)$ out of $2 \sin (x) \cos (x) + 13 \cos (x)$ .

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Step 3.1

Factor $\cos (x)$ out of $2 \sin (x) \cos (x)$ .

$\cos (x) (2 \sin (x)) + 13 \cos (x) = 0$

Step 3.2

Factor $\cos (x)$ out of $13 \cos (x)$ .

$\cos (x) (2 \sin (x)) + \cos (x) \cdot 13 = 0$

Step 3.3

Factor $\cos (x)$ out of $\cos (x) (2 \sin (x)) + \cos (x) \cdot 13$ .

$\cos (x) (2 \sin (x) + 13) = 0$

Step 4

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

$\cos (x) = 0$

$2 \sin (x) + 13 = 0$

Step 5

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

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Step 5.1

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

$\cos (x) = 0$

Step 5.2

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

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Step 5.2.1

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

$x = \arccos (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 5.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{2}$

Step 5.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 5.2.4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 5.2.4.2

Combine fractions.

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Step 5.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$x = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 5.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 5.2.4.3

Simplify the numerator.

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Step 5.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 5.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 5.2.5

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

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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 $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 6

Set $2 \sin (x) + 13$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 \sin (x) + 13$ equal to $0$ .

$2 \sin (x) + 13 = 0$

Step 6.2

Solve $2 \sin (x) + 13 = 0$ for $x$ .

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Step 6.2.1

Subtract $13$ from both sides of the equation.

$2 \sin (x) = - 13$

Step 6.2.2

Divide each term in $2 \sin (x) = - 13$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 \sin (x) = - 13$ by $2$ .

$\frac{2 \sin (x)}{2} = \frac{- 13}{2}$

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.

$\frac{2 \sin (x)}{2} = \frac{- 13}{2}$

Step 6.2.2.2.1.2

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

$\sin (x) = \frac{- 13}{2}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Move the negative in front of the fraction.

$\sin (x) = - \frac{13}{2}$

Step 6.2.3

The range of sine is $- 1 \leq y \leq 1$ . Since $- \frac{13}{2}$ does not fall in this range, there is no solution.

No solution

Step 7

The final solution is all the values that make $\cos (x) (2 \sin (x) + 13) = 0$ true.

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 8

Consolidate the answers.

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