Trigonometry Examples

2 {sin}^{2} (x) - sin (x) = 0

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

Step 1

Factor the left side of the equation.

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Step 1.1

Let

u = sin (x)

$u = \sin (x)$ . Substitute

u

$u$ for all occurrences of

sin (x)

$\sin (x)$ .

2 u^{2} - u = 0

$2 u^{2} - u = 0$

Step 1.2

Factor

u

$u$ out of

2 u^{2} - u

$2 u^{2} - u$ .

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Step 1.2.1

Factor

u

$u$ out of

2 u^{2}

$2 u^{2}$ .

u (2 u) - u = 0

$u (2 u) - u = 0$

Step 1.2.2

Factor

u

$u$ out of

- u

$- u$ .

u (2 u) + u \cdot - 1 = 0

$u (2 u) + u \cdot - 1 = 0$

Step 1.2.3

Factor

u

$u$ out of

u (2 u) + u \cdot - 1

$u (2 u) + u \cdot - 1$ .

u (2 u - 1) = 0

$u (2 u - 1) = 0$

u (2 u - 1) = 0

$u (2 u - 1) = 0$

Step 1.3

Replace all occurrences of

u

$u$ with

sin (x)

$\sin (x)$ .

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

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

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

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

Step 2

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 sin (x) - 1 = 0

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

Step 3

Set

sin (x)

$\sin (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

sin (x)

$\sin (x)$ equal to

0

$0$ .

sin (x) = 0

$\sin (x) = 0$

Step 3.2

Solve

sin (x) = 0

$\sin (x) = 0$ for

x

$x$ .

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Step 3.2.1

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 3.2.2

Simplify the right side.

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Step 3.2.2.1

The exact value of

arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

Step 3.2.3

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 3.2.4

Subtract

0

$0$ from

π

$π$ .

x = π

$x = π$

Step 3.2.5

Find the period of

sin (x)

$\sin (x)$ .

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Step 3.2.5.1

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 3.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 3.2.5.3

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 3.2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 3.2.6

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

$n$

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

$n$

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

$n$

Step 4

Set

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

$0$ and solve for

$x$ .

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Step 4.1

Set

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

$0$ .

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

Step 4.2

Solve

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

$x$ .

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Step 4.2.1

Add

$1$ to both sides of the equation.

$2 \sin (x) = 1$

Step 4.2.2

Divide each term in

$2 \sin (x) = 1$ by

$2$ and simplify.

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Step 4.2.2.1

Divide each term in

$2 \sin (x) = 1$ by

$2$ .

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of

$2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

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

Step 4.2.2.2.1.2

Divide

$\sin (x)$ by

$1$ .

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

Step 4.2.3

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

$x$ from inside the sine.

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

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

The exact value of

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

$\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 4.2.5

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

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

$x = π - \frac{π}{6}$

Step 4.2.6

Simplify

$π - \frac{π}{6}$ .

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Step 4.2.6.1

To write

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

$\frac{6}{6}$ .

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

Step 4.2.6.2

Combine fractions.

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Step 4.2.6.2.1

Combine

$π$ and

$\frac{6}{6}$ .

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

Step 4.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 4.2.6.3

Simplify the numerator.

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Step 4.2.6.3.1

Move

$6$ to the left of

$π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 4.2.6.3.2

Subtract

$π$ from

$6 π$ .

$x = \frac{5 π}{6}$

Step 4.2.7

Find the period of

$\sin (x)$ .

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Step 4.2.7.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2.7.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.7.3

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

$0$ and

$1$ is

$1$ .

$\frac{2 π}{1}$

Step 4.2.7.4

Divide

$2 π$ by

$1$ .

$2 π$

Step 4.2.8

The period of the

$\sin (x)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$

Step 5

The final solution is all the values that make

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

$x = 2 π n, π + 2 π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$

Step 6

Consolidate

$2 π n$ and

$π + 2 π n$ to

$π n$ .

$x = π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$