Precalculus Examples

\cos (x) = - \cos (x) + \sqrt{3}

$\cos (x) = - \cos (x) + \sqrt{3}$

Step 1

Move all terms containing

\cos (x)

$\cos (x)$ to the left side of the equation.

Tap for more steps...

Step 1.1

Add

\cos (x)

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

\cos (x) + \cos (x) = \sqrt{3}

$\cos (x) + \cos (x) = \sqrt{3}$

Step 1.2

Add

\cos (x)

$\cos (x)$ and

\cos (x)

$\cos (x)$ .

2 \cos (x) = \sqrt{3}

$2 \cos (x) = \sqrt{3}$

2 \cos (x) = \sqrt{3}

$2 \cos (x) = \sqrt{3}$

Step 2

Divide each term in

2 \cos (x) = \sqrt{3}

$2 \cos (x) = \sqrt{3}$ by

2

$2$ and simplify.

Tap for more steps...

Step 2.1

Divide each term in

2 \cos (x) = \sqrt{3}

$2 \cos (x) = \sqrt{3}$ by

2

$2$ .

\frac{2 \cos (x)}{2} = \frac{\sqrt{3}}{2}

$\frac{2 \cos (x)}{2} = \frac{\sqrt{3}}{2}$

Step 2.2

Simplify the left side.

Tap for more steps...

Step 2.2.1

Cancel the common factor of

2

$2$ .

Tap for more steps...

Step 2.2.1.1

Cancel the common factor.

\frac{2 \cos (x)}{2} = \frac{\sqrt{3}}{2}

$\frac{2 \cos (x)}{2} = \frac{\sqrt{3}}{2}$

Step 2.2.1.2

Divide

\cos (x)

$\cos (x)$ by

1

$1$ .

\cos (x) = \frac{\sqrt{3}}{2}

$\cos (x) = \frac{\sqrt{3}}{2}$

\cos (x) = \frac{\sqrt{3}}{2}

$\cos (x) = \frac{\sqrt{3}}{2}$

\cos (x) = \frac{\sqrt{3}}{2}

$\cos (x) = \frac{\sqrt{3}}{2}$

\cos (x) = \frac{\sqrt{3}}{2}

$\cos (x) = \frac{\sqrt{3}}{2}$

Step 3

Take the inverse cosine of both sides of the equation to extract

x

$x$ from inside the cosine.

x = \arccos (\frac{\sqrt{3}}{2})

$x = \arccos (\frac{\sqrt{3}}{2})$

Step 4

Simplify the right side.

Tap for more steps...

Step 4.1

The exact value of

\arccos (\frac{\sqrt{3}}{2})

$\arccos (\frac{\sqrt{3}}{2})$ is

\frac{π}{6}

$\frac{π}{6}$ .

x = \frac{π}{6}

$x = \frac{π}{6}$

x = \frac{π}{6}

$x = \frac{π}{6}$

Step 5

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

2 π

$2 π$ to find the solution in the fourth quadrant.

x = 2 π - \frac{π}{6}

$x = 2 π - \frac{π}{6}$

Step 6

Simplify

2 π - \frac{π}{6}

$2 π - \frac{π}{6}$ .

Tap for more steps...

Step 6.1

To write

2 π

$2 π$ as a fraction with a common denominator, multiply by

\frac{6}{6}

$\frac{6}{6}$ .

x = 2 π \cdot \frac{6}{6} - \frac{π}{6}

$x = 2 π \cdot \frac{6}{6} - \frac{π}{6}$

Step 6.2

Combine fractions.

Tap for more steps...

Step 6.2.1

Combine

2 π

$2 π$ and

\frac{6}{6}

$\frac{6}{6}$ .

x = \frac{2 π \cdot 6}{6} - \frac{π}{6}

$x = \frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 6.2.2

Combine the numerators over the common denominator.

x = \frac{2 π \cdot 6 - π}{6}

$x = \frac{2 π \cdot 6 - π}{6}$

x = \frac{2 π \cdot 6 - π}{6}

$x = \frac{2 π \cdot 6 - π}{6}$

Step 6.3

Simplify the numerator.

Tap for more steps...

Step 6.3.1

Multiply

6

$6$ by

2

$2$ .

x = \frac{12 π - π}{6}

$x = \frac{12 π - π}{6}$

Step 6.3.2

Subtract

π

$π$ from

12 π

$12 π$ .

x = \frac{11 π}{6}

$x = \frac{11 π}{6}$

x = \frac{11 π}{6}

$x = \frac{11 π}{6}$

x = \frac{11 π}{6}

$x = \frac{11 π}{6}$

Step 7

Find the period of

\cos (x)

$\cos (x)$ .

Tap for more steps...

Step 7.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 7.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 7.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 8

The period of the

\cos (x)

$\cos (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n

$x = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer

n

$n$

Enter YOUR Problem