Trigonometry Examples

$\cos (2 x) = \cos (x)$

Step 1

Subtract

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

$\cos (2 x) - \cos (x) = 0$

Step 2

Use the double-angle identity to transform

$\cos (2 x)$ to

$2 \cos^{2} (x) - 1$ .

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

Step 3

Factor by grouping.

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Step 3.1

Reorder terms.

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

Step 3.2

For a polynomial of the form

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

$a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is

$b = - 1$ .

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Step 3.2.1

Factor

$- 1$ out of

$- \cos (x)$ .

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

Step 3.2.2

Rewrite

$- 1$ as

$1$ plus

$- 2$

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

Step 3.2.3

Apply the distributive property.

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

Step 3.2.4

Multiply

$\cos (x)$ by

$1$ .

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

Step 3.3

Factor out the greatest common factor from each group.

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Step 3.3.1

Group the first two terms and the last two terms.

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

Step 3.3.2

Factor out the greatest common factor (GCF) from each group.

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

Step 3.4

Factor the polynomial by factoring out the greatest common factor,

$2 \cos (x) + 1$ .

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

Step 4

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

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

$\cos (x) - 1 = 0$

Step 5

Set

$2 \cos (x) + 1$ equal to

$0$ and solve for

$x$ .

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Step 5.1

Set

$2 \cos (x) + 1$ equal to

$0$ .

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

Step 5.2

Solve

$2 \cos (x) + 1 = 0$ for

$x$ .

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Step 5.2.1

Subtract

$1$ from both sides of the equation.

$2 \cos (x) = - 1$

Step 5.2.2

Divide each term in

$2 \cos (x) = - 1$ by

$2$ and simplify.

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Step 5.2.2.1

Divide each term in

$2 \cos (x) = - 1$ by

$2$ .

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

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of

$2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.2.1.2

Divide

$\cos (x)$ by

$1$ .

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

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

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

Step 5.2.3

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

$x$ from inside the cosine.

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

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

The exact value of

$\arccos (- \frac{1}{2})$ is

$\frac{2 π}{3}$ .

$x = \frac{2 π}{3}$

Step 5.2.5

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from

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

$x = 2 π - \frac{2 π}{3}$

Step 5.2.6

Simplify

$2 π - \frac{2 π}{3}$ .

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Step 5.2.6.1

To write

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

$\frac{3}{3}$ .

$x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Step 5.2.6.2

Combine fractions.

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Step 5.2.6.2.1

Combine

$2 π$ and

$\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Step 5.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 5.2.6.3

Simplify the numerator.

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Step 5.2.6.3.1

Multiply

$3$ by

$2$ .

$x = \frac{6 π - 2 π}{3}$

Step 5.2.6.3.2

Subtract

$2 π$ from

$6 π$ .

$x = \frac{4 π}{3}$

Step 5.2.7

Find the period of

$\cos (x)$ .

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Step 5.2.7.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 5.2.7.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 5.2.7.3

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

$0$ and

$1$ is

$1$ .

$\frac{2 π}{1}$

Step 5.2.7.4

Divide

$2 π$ by

$1$ .

$2 π$

Step 5.2.8

The period of the

$\cos (x)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer

$n$

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer

$n$

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer

$n$

Step 6

Set

$\cos (x) - 1$ equal to

$0$ and solve for

$x$ .

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Step 6.1

Set

$\cos (x) - 1$ equal to

$0$ .

$\cos (x) - 1 = 0$

Step 6.2

Solve

$\cos (x) - 1 = 0$ for

$x$ .

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Step 6.2.1

Add

$1$ to both sides of the equation.

$\cos (x) = 1$

Step 6.2.2

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

$x$ from inside the cosine.

$x = \arccos (1)$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

The exact value of

$\arccos (1)$ is

$0$ .

$x = 0$

Step 6.2.4

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.

$x = 2 π - 0$

Step 6.2.5

Subtract

$0$ from

$2 π$ .

$x = 2 π$

Step 6.2.6

Find the period of

$\cos (x)$ .

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Step 6.2.6.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 6.2.6.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 6.2.6.3

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

$0$ and

$1$ is

$1$ .

$\frac{2 π}{1}$

Step 6.2.6.4

Divide

$2 π$ by

$1$ .

$2 π$

Step 6.2.7

The period of the

$\cos (x)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

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

$n$

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

$n$

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

$n$

Step 7

The final solution is all the values that make

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

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, 2 π n, 2 π + 2 π n$ , for any integer

$n$

Step 8

Consolidate the answers.

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

$n$