Trigonometry Examples

sin (2 x) = \sqrt{2} sin (x)

$\sin (2 x) = \sqrt{2} \sin (x)$

Step 1

Subtract

\sqrt{2} sin (x)

$\sqrt{2} \sin (x)$ from both sides of the equation.

sin (2 x) - \sqrt{2} sin (x) = 0

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

Step 2

Apply the sine double-angle identity.

2 sin (x) cos (x) - \sqrt{2} sin (x) = 0

$2 \sin (x) \cos (x) - \sqrt{2} \sin (x) = 0$

Step 3

Factor

sin (x)

$\sin (x)$ out of

2 sin (x) cos (x) - \sqrt{2} sin (x)

$2 \sin (x) \cos (x) - \sqrt{2} \sin (x)$ .

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Step 3.1

Factor

sin (x)

$\sin (x)$ out of

2 sin (x) cos (x)

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

sin (x) (2 cos (x)) - \sqrt{2} sin (x) = 0

$\sin (x) (2 \cos (x)) - \sqrt{2} \sin (x) = 0$

Step 3.2

Factor

sin (x)

$\sin (x)$ out of

- \sqrt{2} sin (x)

$- \sqrt{2} \sin (x)$ .

sin (x) (2 cos (x)) + sin (x) (- \sqrt{2}) = 0

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

Step 3.3

Factor

sin (x)

$\sin (x)$ out of

sin (x) (2 cos (x)) + sin (x) (- \sqrt{2})

$\sin (x) (2 \cos (x)) + \sin (x) (- \sqrt{2})$ .

sin (x) (2 cos (x) - \sqrt{2}) = 0

$\sin (x) (2 \cos (x) - \sqrt{2}) = 0$

sin (x) (2 cos (x) - \sqrt{2}) = 0

$\sin (x) (2 \cos (x) - \sqrt{2}) = 0$

Step 4

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

0

$0$ , the entire expression will be equal to

0

$0$ .

sin (x) = 0

$\sin (x) = 0$

2 cos (x) - \sqrt{2} = 0

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

Step 5

Set

sin (x)

$\sin (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

sin (x)

$\sin (x)$ equal to

0

$0$ .

sin (x) = 0

$\sin (x) = 0$

Step 5.2

Solve

sin (x) = 0

$\sin (x) = 0$ for

x

$x$ .

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Step 5.2.1

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

x

$x$ from inside the sine.

x = arcsin (0)

$x = \arcsin (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of

arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x = 0

$x = 0$

x = 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

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

x = π - 0

$x = π - 0$

Step 5.2.4

Subtract

0

$0$ from

π

$π$ .

x = π

$x = π$

Step 5.2.5

Find the period of

sin (x)

$\sin (x)$ .

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Step 5.2.5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 5.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 5.2.5.3

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 5.2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 5.2.6

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 = 2 π n, π + 2 π n

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

n

$n$

x = 2 π n, π + 2 π n

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

n

$n$

x = 2 π n, π + 2 π n

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

n

$n$

Step 6

Set

2 cos (x) - \sqrt{2}

$2 \cos (x) - \sqrt{2}$ equal to

0

$0$ and solve for

x

$x$ .

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Step 6.1

Set

2 cos (x) - \sqrt{2}

$2 \cos (x) - \sqrt{2}$ equal to

0

$0$ .

2 cos (x) - \sqrt{2} = 0

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

Step 6.2

Solve

2 cos (x) - \sqrt{2} = 0

$2 \cos (x) - \sqrt{2} = 0$ for

x

$x$ .

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Step 6.2.1

Add

\sqrt{2}

$\sqrt{2}$ to both sides of the equation.

2 cos (x) = \sqrt{2}

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

Step 6.2.2

Divide each term in

2 cos (x) = \sqrt{2}

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

2

$2$ and simplify.

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Step 6.2.2.1

Divide each term in

2 cos (x) = \sqrt{2}

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

2

$2$ .

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

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

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

$2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

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

Step 6.2.2.2.1.2

Divide

$\cos (x)$ by

$1$ .

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

Step 6.2.3

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

$x$ from inside the cosine.

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

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

The exact value of

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

$\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 6.2.5

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.

$x = 2 π - \frac{π}{4}$

Step 6.2.6

Simplify

$2 π - \frac{π}{4}$ .

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Step 6.2.6.1

To write

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

$\frac{4}{4}$ .

$x = 2 π \cdot \frac{4}{4} - \frac{π}{4}$

Step 6.2.6.2

Combine fractions.

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Step 6.2.6.2.1

Combine

$2 π$ and

$\frac{4}{4}$ .

$x = \frac{2 π \cdot 4}{4} - \frac{π}{4}$

Step 6.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 4 - π}{4}$

Step 6.2.6.3

Simplify the numerator.

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Step 6.2.6.3.1

Multiply

$4$ by

$2$ .

$x = \frac{8 π - π}{4}$

Step 6.2.6.3.2

Subtract

$π$ from

$8 π$ .

$x = \frac{7 π}{4}$

Step 6.2.7

Find the period of

$\cos (x)$ .

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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

$\cos (x)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

$x = \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer

$n$

$x = \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer

$n$

$x = \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer

$n$

Step 7

The final solution is all the values that make

$\sin (x) (2 \cos (x) - \sqrt{2}) = 0$ true.

$x = 2 π n, π + 2 π n, \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer

$n$

Step 8

Consolidate

$2 π n$ and

$π + 2 π n$ to

$π n$ .

$x = π n, \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer

$n$