Trigonometry Examples

2 {sin}^{2} (x) - sin (x) - 1 = 0

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

Step 1

Factor by grouping.

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Step 1.1

For a polynomial of the form

a x^{2} + b x + c

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

a \cdot c = 2 \cdot - 1 = - 2

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

b = - 1

$b = - 1$ .

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Step 1.1.1

Factor

- 1

$- 1$ out of

- sin (x)

$- \sin (x)$ .

2 {sin}^{2} (x) - sin (x) - 1 = 0

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

Step 1.1.2

Rewrite

- 1

$- 1$ as

1

$1$ plus

- 2

$- 2$

2 {sin}^{2} (x) + (1 - 2) sin (x) - 1 = 0

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

Step 1.1.3

Apply the distributive property.

2 {sin}^{2} (x) + 1 sin (x) - 2 sin (x) - 1 = 0

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

Step 1.1.4

Multiply

sin (x)

$\sin (x)$ by

1

$1$ .

2 {sin}^{2} (x) + sin (x) - 2 sin (x) - 1 = 0

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

2 {sin}^{2} (x) + sin (x) - 2 sin (x) - 1 = 0

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

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

2 {sin}^{2} (x) + sin (x) - 2 sin (x) - 1 = 0

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

Step 1.2.2

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

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

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

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

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

Step 1.3

Factor the polynomial by factoring out the greatest common factor,

2 sin (x) + 1

$2 \sin (x) + 1$ .

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

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

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

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

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

$0$ .

2 sin (x) + 1 = 0

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

sin (x) - 1 = 0

$\sin (x) - 1 = 0$

Step 3

Set

2 sin (x) + 1

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

0

$0$ and solve for

x

$x$ .

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Step 3.1

Set

2 sin (x) + 1

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

0

$0$ .

2 sin (x) + 1 = 0

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

Step 3.2

Solve

2 sin (x) + 1 = 0

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

x

$x$ .

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Step 3.2.1

Subtract

1

$1$ from both sides of the equation.

2 sin (x) = - 1

$2 \sin (x) = - 1$

Step 3.2.2

Divide each term in

2 sin (x) = - 1

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

2

$2$ and simplify.

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Step 3.2.2.1

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 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

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

Step 3.2.2.2.1.2

Divide

$\sin (x)$ by

$1$ .

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

Step 3.2.2.3

Simplify the right side.

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Step 3.2.2.3.1

Move the negative in front of the fraction.

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

Step 3.2.3

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

$x$ from inside the sine.

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

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

The exact value of

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

$- \frac{π}{6}$ .

$x = - \frac{π}{6}$

Step 3.2.5

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from

$2 π$ , to find a reference angle. Next, add this reference angle to

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

$x = 2 π + \frac{π}{6} + π$

Step 3.2.6

Simplify the expression to find the second solution.

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Step 3.2.6.1

Subtract

$2 π$ from

$2 π + \frac{π}{6} + π$ .

$x = 2 π + \frac{π}{6} + π - 2 π$

Step 3.2.6.2

The resulting angle of

$\frac{7 π}{6}$ is positive, less than

$2 π$ , and coterminal with

$2 π + \frac{π}{6} + π$ .

$x = \frac{7 π}{6}$

Step 3.2.7

Find the period of

$\sin (x)$ .

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Step 3.2.7.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.2.7.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.2.7.3

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

$0$ and

$1$ is

$1$ .

$\frac{2 π}{1}$

Step 3.2.7.4

Divide

$2 π$ by

$1$ .

$2 π$

Step 3.2.8

Add

$2 π$ to every negative angle to get positive angles.

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Step 3.2.8.1

Add

$2 π$ to

$- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

Step 3.2.8.2

To write

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

$\frac{6}{6}$ .

$2 π \cdot \frac{6}{6} - \frac{π}{6}$

Step 3.2.8.3

Combine fractions.

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Step 3.2.8.3.1

Combine

$2 π$ and

$\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 3.2.8.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 3.2.8.4

Simplify the numerator.

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Step 3.2.8.4.1

Multiply

$6$ by

$2$ .

$\frac{12 π - π}{6}$

Step 3.2.8.4.2

Subtract

$π$ from

$12 π$ .

$\frac{11 π}{6}$

Step 3.2.8.5

List the new angles.

$x = \frac{11 π}{6}$

Step 3.2.9

The period of the

$\sin (x)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer

$n$

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer

$n$

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer

$n$

Step 4

Set

$\sin (x) - 1$ equal to

$0$ and solve for

$x$ .

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Step 4.1

Set

$\sin (x) - 1$ equal to

$0$ .

$\sin (x) - 1 = 0$

Step 4.2

Solve

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

$x$ .

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Step 4.2.1

Add

$1$ to both sides of the equation.

$\sin (x) = 1$

Step 4.2.2

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

$x$ from inside the sine.

$x = \arcsin (1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

The exact value of

$\arcsin (1)$ is

$\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 4.2.4

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.

$x = π - \frac{π}{2}$

Step 4.2.5

Simplify

$π - \frac{π}{2}$ .

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Step 4.2.5.1

To write

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

$\frac{2}{2}$ .

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

Step 4.2.5.2

Combine fractions.

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Step 4.2.5.2.1

Combine

$π$ and

$\frac{2}{2}$ .

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

Step 4.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 4.2.5.3

Simplify the numerator.

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Step 4.2.5.3.1

Move

$2$ to the left of

$π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 4.2.5.3.2

Subtract

$π$ from

$2 π$ .

$x = \frac{π}{2}$

Step 4.2.6

Find the period of

$\sin (x)$ .

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Step 4.2.6.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2.6.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.6.3

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

$0$ and

$1$ is

$1$ .

$\frac{2 π}{1}$

Step 4.2.6.4

Divide

$2 π$ by

$1$ .

$2 π$

Step 4.2.7

The period of the

$\sin (x)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

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

$n$

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

$n$

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

$n$

Step 5

The final solution is all the values that make

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

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n, \frac{π}{2} + 2 π n$ , for any integer

$n$

Step 6

Consolidate the answers.

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

$n$