Trigonometry Examples

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

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

(0, 2 π)

$(0, 2 π)$

Step 1

Apply the sine double-angle identity.

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

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

Step 2

Factor

2 \cos (x)

$2 \cos (x)$ out of

2 \sin (x) \cos (x) - 2 \cos (x)

$2 \sin (x) \cos (x) - 2 \cos (x)$ .

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Step 2.1

Factor

2 \cos (x)

$2 \cos (x)$ out of

2 \sin (x) \cos (x)

$2 \sin (x) \cos (x)$ .

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

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

Step 2.2

Factor

2 \cos (x)

$2 \cos (x)$ out of

- 2 \cos (x)

$- 2 \cos (x)$ .

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

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

Step 2.3

Factor

2 \cos (x)

$2 \cos (x)$ out of

2 \cos (x) \sin (x) + 2 \cos (x) \cdot - 1

$2 \cos (x) \sin (x) + 2 \cos (x) \cdot - 1$ .

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

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

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

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

Step 3

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

0

$0$ , the entire expression will be equal to

0

$0$ .

\cos (x) = 0

$\cos (x) = 0$

\sin (x) - 1 = 0

$\sin (x) - 1 = 0$

Step 4

Set

\cos (x)

$\cos (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 4.1

Set

\cos (x)

$\cos (x)$ equal to

0

$0$ .

\cos (x) = 0

$\cos (x) = 0$

Step 4.2

Solve

\cos (x) = 0

$\cos (x) = 0$ for

x

$x$ .

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Step 4.2.1

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

x

$x$ from inside the cosine.

x = \arccos (0)

$x = \arccos (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

The exact value of

\arccos (0)

$\arccos (0)$ is

\frac{π}{2}

$\frac{π}{2}$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 4.2.3

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

2 π

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

x = 2 π - \frac{π}{2}

$x = 2 π - \frac{π}{2}$

Step 4.2.4

Simplify

2 π - \frac{π}{2}

$2 π - \frac{π}{2}$ .

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Step 4.2.4.1

To write

2 π

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

\frac{2}{2}

$\frac{2}{2}$ .

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

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

Step 4.2.4.2

Combine fractions.

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Step 4.2.4.2.1

Combine

2 π

$2 π$ and

\frac{2}{2}

$\frac{2}{2}$ .

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

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

Step 4.2.4.2.2

Combine the numerators over the common denominator.

x = \frac{2 π \cdot 2 - π}{2}

$x = \frac{2 π \cdot 2 - π}{2}$

x = \frac{2 π \cdot 2 - π}{2}

$x = \frac{2 π \cdot 2 - π}{2}$

Step 4.2.4.3

Simplify the numerator.

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Step 4.2.4.3.1

Multiply

2

$2$ by

2

$2$ .

x = \frac{4 π - π}{2}

$x = \frac{4 π - π}{2}$

Step 4.2.4.3.2

Subtract

π

$π$ from

4 π

$4 π$ .

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

Step 4.2.5

Find the period of

\cos (x)

$\cos (x)$ .

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Step 4.2.5.1

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 4.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 4.2.5.3

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 4.2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 4.2.6

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}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

Step 5

Set

\sin (x) - 1

$\sin (x) - 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

\sin (x) - 1

$\sin (x) - 1$ equal to

0

$0$ .

\sin (x) - 1 = 0

$\sin (x) - 1 = 0$

Step 5.2

Solve

\sin (x) - 1 = 0

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

x

$x$ .

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Step 5.2.1

Add

1

$1$ to both sides of the equation.

\sin (x) = 1

$\sin (x) = 1$

Step 5.2.2

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

x

$x$ from inside the sine.

x = \arcsin (1)

$x = \arcsin (1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

The exact value of

\arcsin (1)

$\arcsin (1)$ is

\frac{π}{2}

$\frac{π}{2}$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 5.2.4

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

π

$π$ to find the solution in the second quadrant.

x = π - \frac{π}{2}

$x = π - \frac{π}{2}$

Step 5.2.5

Simplify

π - \frac{π}{2}

$π - \frac{π}{2}$ .

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Step 5.2.5.1

To write

π

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

\frac{2}{2}

$\frac{2}{2}$ .

x = π \cdot \frac{2}{2} - \frac{π}{2}

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

Step 5.2.5.2

Combine fractions.

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Step 5.2.5.2.1

Combine

π

$π$ and

\frac{2}{2}

$\frac{2}{2}$ .

x = \frac{π \cdot 2}{2} - \frac{π}{2}

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

Step 5.2.5.2.2

Combine the numerators over the common denominator.

x = \frac{π \cdot 2 - π}{2}

$x = \frac{π \cdot 2 - π}{2}$

x = \frac{π \cdot 2 - π}{2}

$x = \frac{π \cdot 2 - π}{2}$

Step 5.2.5.3

Simplify the numerator.

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Step 5.2.5.3.1

Move

2

$2$ to the left of

π

$π$ .

x = \frac{2 \cdot π - π}{2}

$x = \frac{2 \cdot π - π}{2}$

Step 5.2.5.3.2

Subtract

π

$π$ from

2 π

$2 π$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 5.2.6

Find the period of

\sin (x)

$\sin (x)$ .

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Step 5.2.6.1

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 5.2.6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 5.2.6.3

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 5.2.6.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 5.2.7

The period of the

\sin (x)

$\sin (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = \frac{π}{2} + 2 π n

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

n

$n$

x = \frac{π}{2} + 2 π n

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

n

$n$

x = \frac{π}{2} + 2 π n

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

n

$n$

Step 6

The final solution is all the values that make

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

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

x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

Step 7

Consolidate the answers.

x = \frac{π}{2} + π n

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

n

$n$

Step 8

Find the values of

n

$n$ that produce a value within the interval

(0, 2 π)

$(0, 2 π)$ .

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Step 8.1

Plug in

0

$0$ for

n

$n$ and simplify to see if the solution is contained in

(0, 2 π)

$(0, 2 π)$ .

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Step 8.1.1

Plug in

0

$0$ for

n

$n$ .

\frac{π}{2} + π (0)

$\frac{π}{2} + π (0)$

Step 8.1.2

Simplify.

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Step 8.1.2.1

Multiply

π

$π$ by

0

$0$ .

\frac{π}{2} + 0

$\frac{π}{2} + 0$

Step 8.1.2.2

Add

\frac{π}{2}

$\frac{π}{2}$ and

0

$0$ .

\frac{π}{2}

$\frac{π}{2}$

\frac{π}{2}

$\frac{π}{2}$

Step 8.1.3

The interval

(0, 2 π)

$(0, 2 π)$ contains

\frac{π}{2}

$\frac{π}{2}$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 8.2

Plug in

1

$1$ for

n

$n$ and simplify to see if the solution is contained in

(0, 2 π)

$(0, 2 π)$ .

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Step 8.2.1

Plug in

1

$1$ for

n

$n$ .

\frac{π}{2} + π (1)

$\frac{π}{2} + π (1)$

Step 8.2.2

Simplify.

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Step 8.2.2.1

Multiply

π

$π$ by

1

$1$ .

\frac{π}{2} + π

$\frac{π}{2} + π$

Step 8.2.2.2

To write

π

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

\frac{2}{2}

$\frac{2}{2}$ .

\frac{π}{2} + π \cdot \frac{2}{2}

$\frac{π}{2} + π \cdot \frac{2}{2}$

Step 8.2.2.3

Combine fractions.

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Step 8.2.2.3.1

Combine

π

$π$ and

\frac{2}{2}

$\frac{2}{2}$ .

\frac{π}{2} + \frac{π \cdot 2}{2}

$\frac{π}{2} + \frac{π \cdot 2}{2}$

Step 8.2.2.3.2

Combine the numerators over the common denominator.

\frac{π + π \cdot 2}{2}

$\frac{π + π \cdot 2}{2}$

\frac{π + π \cdot 2}{2}

$\frac{π + π \cdot 2}{2}$

Step 8.2.2.4

Simplify the numerator.

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Step 8.2.2.4.1

Move

2

$2$ to the left of

π

$π$ .

\frac{π + 2 \cdot π}{2}

$\frac{π + 2 \cdot π}{2}$

Step 8.2.2.4.2

Add

π

$π$ and

2 π

$2 π$ .

\frac{3 π}{2}

$\frac{3 π}{2}$

\frac{3 π}{2}

$\frac{3 π}{2}$

\frac{3 π}{2}

$\frac{3 π}{2}$

Step 8.2.3

The interval

(0, 2 π)

$(0, 2 π)$ contains

\frac{3 π}{2}

$\frac{3 π}{2}$ .

x = \frac{π}{2}, \frac{3 π}{2}

$x = \frac{π}{2}, \frac{3 π}{2}$

x = \frac{π}{2}, \frac{3 π}{2}

$x = \frac{π}{2}, \frac{3 π}{2}$

x = \frac{π}{2}, \frac{3 π}{2}

$x = \frac{π}{2}, \frac{3 π}{2}$