Trigonometry Examples

6 {sin}^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

$6 \sin^{2} (θ) - 17 \sin (θ) + 14 = - 4 \sin (θ) + 9$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Add

4 sin (θ)

$4 \sin (θ)$ to both sides of the equation.

6 {sin}^{2} (θ) - 17 sin (θ) + 14 + 4 sin (θ) = 9

$6 \sin^{2} (θ) - 17 \sin (θ) + 14 + 4 \sin (θ) = 9$

Step 1.2

Subtract

9

$9$ from both sides of the equation.

6 {sin}^{2} (θ) - 17 sin (θ) + 14 + 4 sin (θ) - 9 = 0

$6 \sin^{2} (θ) - 17 \sin (θ) + 14 + 4 \sin (θ) - 9 = 0$

6 {sin}^{2} (θ) - 17 sin (θ) + 14 + 4 sin (θ) - 9 = 0

$6 \sin^{2} (θ) - 17 \sin (θ) + 14 + 4 \sin (θ) - 9 = 0$

Step 2

Simplify the left side of the equation.

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Step 2.1

Add

- 17 sin (θ)

$- 17 \sin (θ)$ and

4 sin (θ)

$4 \sin (θ)$ .

6 {sin}^{2} (θ) + 14 - 13 sin (θ) - 9 = 0

$6 \sin^{2} (θ) + 14 - 13 \sin (θ) - 9 = 0$

Step 2.2

Subtract

9

$9$ from

14

$14$ .

6 {sin}^{2} (θ) + 5 - 13 sin (θ) = 0

$6 \sin^{2} (θ) + 5 - 13 \sin (θ) = 0$

6 {sin}^{2} (θ) + 5 - 13 sin (θ) = 0

$6 \sin^{2} (θ) + 5 - 13 \sin (θ) = 0$

Step 3

Factor by grouping.

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Step 3.1

Reorder terms.

6 {sin}^{2} (θ) - 13 sin (θ) + 5 = 0

$6 \sin^{2} (θ) - 13 \sin (θ) + 5 = 0$

Step 3.2

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 = 6 \cdot 5 = 30

$a \cdot c = 6 \cdot 5 = 30$ and whose sum is

b = - 13

$b = - 13$ .

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Step 3.2.1

Factor

- 13

$- 13$ out of

- 13 sin (θ)

$- 13 \sin (θ)$ .

6 {sin}^{2} (θ) - 13 sin (θ) + 5 = 0

$6 \sin^{2} (θ) - 13 \sin (θ) + 5 = 0$

Step 3.2.2

Rewrite

- 13

$- 13$ as

- 3

$- 3$ plus

- 10

$- 10$

6 {sin}^{2} (θ) + (- 3 - 10) sin (θ) + 5 = 0

$6 \sin^{2} (θ) + (- 3 - 10) \sin (θ) + 5 = 0$

Step 3.2.3

Apply the distributive property.

6 {sin}^{2} (θ) - 3 sin (θ) - 10 sin (θ) + 5 = 0

$6 \sin^{2} (θ) - 3 \sin (θ) - 10 \sin (θ) + 5 = 0$

6 {sin}^{2} (θ) - 3 sin (θ) - 10 sin (θ) + 5 = 0

$6 \sin^{2} (θ) - 3 \sin (θ) - 10 \sin (θ) + 5 = 0$

Step 3.3

Factor out the greatest common factor from each group.

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Step 3.3.1

Group the first two terms and the last two terms.

6 {sin}^{2} (θ) - 3 sin (θ) - 10 sin (θ) + 5 = 0

$6 \sin^{2} (θ) - 3 \sin (θ) - 10 \sin (θ) + 5 = 0$

Step 3.3.2

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

3 sin (θ) (2 sin (θ) - 1) - 5 (2 sin (θ) - 1) = 0

$3 \sin (θ) (2 \sin (θ) - 1) - 5 (2 \sin (θ) - 1) = 0$

3 sin (θ) (2 sin (θ) - 1) - 5 (2 sin (θ) - 1) = 0

$3 \sin (θ) (2 \sin (θ) - 1) - 5 (2 \sin (θ) - 1) = 0$

Step 3.4

Factor the polynomial by factoring out the greatest common factor,

2 sin (θ) - 1

$2 \sin (θ) - 1$ .

(2 sin (θ) - 1) (3 sin (θ) - 5) = 0

$(2 \sin (θ) - 1) (3 \sin (θ) - 5) = 0$

(2 sin (θ) - 1) (3 sin (θ) - 5) = 0

$(2 \sin (θ) - 1) (3 \sin (θ) - 5) = 0$

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

$0$ .

2 sin (θ) - 1 = 0

$2 \sin (θ) - 1 = 0$

3 sin (θ) - 5 = 0

$3 \sin (θ) - 5 = 0$

Step 5

Set

2 sin (θ) - 1

$2 \sin (θ) - 1$ equal to

0

$0$ and solve for

θ

$θ$ .

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Step 5.1

Set

2 sin (θ) - 1

$2 \sin (θ) - 1$ equal to

0

$0$ .

2 sin (θ) - 1 = 0

$2 \sin (θ) - 1 = 0$

Step 5.2

Solve

2 sin (θ) - 1 = 0

$2 \sin (θ) - 1 = 0$ for

θ

$θ$ .

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Step 5.2.1

Add

1

$1$ to both sides of the equation.

2 sin (θ) = 1

$2 \sin (θ) = 1$

Step 5.2.2

Divide each term in

2 sin (θ) = 1

$2 \sin (θ) = 1$ by

2

$2$ and simplify.

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Step 5.2.2.1

Divide each term in

2 sin (θ) = 1

$2 \sin (θ) = 1$ by

2

$2$ .

\frac{2 sin (θ)}{2} = \frac{1}{2}

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

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.

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Step 5.2.2.2.1.2

Divide

$\sin (θ)$ by

$1$ .

$\sin (θ) = \frac{1}{2}$

Step 5.2.3

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

$θ$ from inside the sine.

$θ = \arcsin (\frac{1}{2})$

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

$30$ .

$θ = 30$

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.

$θ = 180 - 30$

Step 5.2.6

Subtract

$30$ from

$180$ .

$θ = 150$

Step 5.2.7

Find the period of

$\sin (θ)$ .

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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

$\sin (θ)$ function is

$360$ so values will repeat every

$360$ degrees in both directions.

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

$n$

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

$n$

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

$n$

Step 6

Set

$3 \sin (θ) - 5$ equal to

$0$ and solve for

$θ$ .

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Step 6.1

Set

$3 \sin (θ) - 5$ equal to

$0$ .

$3 \sin (θ) - 5 = 0$

Step 6.2

Solve

$3 \sin (θ) - 5 = 0$ for

$θ$ .

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Step 6.2.1

Add

$5$ to both sides of the equation.

$3 \sin (θ) = 5$

Step 6.2.2

Divide each term in

$3 \sin (θ) = 5$ by

$3$ and simplify.

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Step 6.2.2.1

Divide each term in

$3 \sin (θ) = 5$ by

$3$ .

$\frac{3 \sin (θ)}{3} = \frac{5}{3}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of

$3$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{3 \sin (θ)}{3} = \frac{5}{3}$

Step 6.2.2.2.1.2

Divide

$\sin (θ)$ by

$1$ .

$\sin (θ) = \frac{5}{3}$

Step 6.2.3

The range of sine is

$- 1 \leq y \leq 1$ . Since

$\frac{5}{3}$ does not fall in this range, there is no solution.

No solution

Step 7

The final solution is all the values that make

$(2 \sin (θ) - 1) (3 \sin (θ) - 5) = 0$ true.

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

$n$