Trigonometry Examples

2 sin (x) cos (x) = cos (x)

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

Step 1

Subtract

cos (x)

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

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

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

Step 2

Factor

cos (x)

$\cos (x)$ out of

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

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

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Step 2.1

Factor

cos (x)

$\cos (x)$ out of

2 sin (x) cos (x)

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

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

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

Step 2.2

Factor

cos (x)

$\cos (x)$ out of

- cos (x)

$- \cos (x)$ .

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

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

Step 2.3

Factor

cos (x)

$\cos (x)$ out of

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

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

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

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

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

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

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

$\cos (x) = 0$

2 sin (x) - 1 = 0

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

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

90

$90$ .

x = 90

$x = 90$

x = 90

$x = 90$

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

360

$360$ to find the solution in the fourth quadrant.

x = 360 - 90

$x = 360 - 90$

Step 4.2.4

Subtract

90

$90$ from

360

$360$ .

x = 270

$x = 270$

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{360}{| b |}

$\frac{360}{| b |}$ .

\frac{360}{| b |}

$\frac{360}{| b |}$

Step 4.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{360}{| 1 |}

$\frac{360}{| 1 |}$

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{360}{1}

$\frac{360}{1}$

Step 4.2.5.4

Divide

360

$360$ by

1

$1$ .

360

$360$

360

$360$

Step 4.2.6

The period of the

cos (x)

$\cos (x)$ function is

360

$360$ so values will repeat every

360

$360$ degrees in both directions.

x = 90 + 360 n, 270 + 360 n

$x = 90 + 360 n, 270 + 360 n$ , for any integer

n

$n$

x = 90 + 360 n, 270 + 360 n

$x = 90 + 360 n, 270 + 360 n$ , for any integer

n

$n$

x = 90 + 360 n, 270 + 360 n

$x = 90 + 360 n, 270 + 360 n$ , for any integer

n

$n$

Step 5

Set

2 sin (x) - 1

$2 \sin (x) - 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

2 sin (x) - 1

$2 \sin (x) - 1$ equal to

0

$0$ .

2 sin (x) - 1 = 0

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

Step 5.2

Solve

2 sin (x) - 1 = 0

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

x

$x$ .

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Step 5.2.1

Add

1

$1$ to both sides of the equation.

2 sin (x) = 1

$2 \sin (x) = 1$

Step 5.2.2

Divide each term in

2 sin (x) = 1

$2 \sin (x) = 1$ by

2

$2$ and simplify.

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Step 5.2.2.1

Divide each term in

2 sin (x) = 1

$2 \sin (x) = 1$ by

2

$2$ .

\frac{2 sin (x)}{2} = \frac{1}{2}

$\frac{2 \sin (x)}{2} = \frac{1}{2}$

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

$2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 \sin (x)}{2} = \frac{1}{2}$

Step 5.2.2.2.1.2

Divide

$\sin (x)$ by

$1$ .

$\sin (x) = \frac{1}{2}$

Step 5.2.3

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

$x$ from inside the sine.

$x = \arcsin (\frac{1}{2})$

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

The exact value of

$\arcsin (\frac{1}{2})$ is

$30$ .

$x = 30$

Step 5.2.5

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

$180$ to find the solution in the second quadrant.

$x = 180 - 30$

Step 5.2.6

Subtract

$30$ from

$180$ .

$x = 150$

Step 5.2.7

Find the period of

$\sin (x)$ .

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Step 5.2.7.1

The period of the function can be calculated using

$\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 5.2.7.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{360}{| 1 |}$

Step 5.2.7.3

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

$0$ and

$1$ is

$1$ .

$\frac{360}{1}$

Step 5.2.7.4

Divide

$360$ by

$1$ .

$360$

Step 5.2.8

The period of the

$\sin (x)$ function is

$360$ so values will repeat every

$360$ degrees in both directions.

$x = 30 + 360 n, 150 + 360 n$ , for any integer

$n$

$x = 30 + 360 n, 150 + 360 n$ , for any integer

$n$

$x = 30 + 360 n, 150 + 360 n$ , for any integer

$n$

Step 6

The final solution is all the values that make

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

$x = 90 + 360 n, 270 + 360 n, 30 + 360 n, 150 + 360 n$ , for any integer

$n$

Step 7

Consolidate

$90 + 360 n$ and

$270 + 360 n$ to

$90 + 180 n$ .

$x = 90 + 180 n, 30 + 360 n, 150 + 360 n$ , for any integer

$n$