Trigonometry Examples

$9 \cos^{2} (θ) - 24 \sin (θ) - 10 = - 8 \sin (θ) + 6$

Step 1

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

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Step 1.1

Add $8 \sin (θ)$ to both sides of the equation.

$9 \cos^{2} (θ) - 24 \sin (θ) - 10 + 8 \sin (θ) = 6$

Step 1.2

Subtract $6$ from both sides of the equation.

$9 \cos^{2} (θ) - 24 \sin (θ) - 10 + 8 \sin (θ) - 6 = 0$

Step 2

Simplify $9 \cos^{2} (θ) - 24 \sin (θ) - 10 + 8 \sin (θ) - 6$ .

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Step 2.1

Add $- 24 \sin (θ)$ and $8 \sin (θ)$ .

$9 \cos^{2} (θ) - 10 - 16 \sin (θ) - 6 = 0$

Step 2.2

Subtract $6$ from $- 10$ .

$9 \cos^{2} (θ) - 16 - 16 \sin (θ) = 0$

Step 3

Replace the $9 \cos^{2} (θ)$ with $9 (1 - \sin^{2} (θ))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$9 (1 - \sin^{2} (θ)) - 16 - 16 \sin (θ) = 0$

Step 4

Simplify each term.

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Step 4.1

Apply the distributive property.

$9 \cdot 1 + 9 (- \sin^{2} (θ)) - 16 - 16 \sin (θ) = 0$

Step 4.2

Multiply $9$ by $1$ .

$9 + 9 (- \sin^{2} (θ)) - 16 - 16 \sin (θ) = 0$

Step 4.3

Multiply $- 1$ by $9$ .

$9 - 9 \sin^{2} (θ) - 16 - 16 \sin (θ) = 0$

Step 5

Subtract $16$ from $9$ .

$- 9 \sin^{2} (θ) - 7 - 16 \sin (θ) = 0$

Step 6

Reorder the polynomial.

$- 9 \sin^{2} (θ) - 16 \sin (θ) - 7 = 0$

Step 7

Substitute $u$ for $\sin (θ)$ .

$- 9 {(u)}^{2} - 16 u - 7 = 0$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $- 1$ out of $- 9 u^{2} - 16 u - 7$ .

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Step 8.1.1

Factor $- 1$ out of $- 9 u^{2}$ .

$- (9 u^{2}) - 16 u - 7 = 0$

Step 8.1.2

Factor $- 1$ out of $- 16 u$ .

$- (9 u^{2}) - (16 u) - 7 = 0$

Step 8.1.3

Rewrite $- 7$ as $- 1 (7)$ .

$- (9 u^{2}) - (16 u) - 1 \cdot 7 = 0$

Step 8.1.4

Factor $- 1$ out of $- (9 u^{2}) - (16 u)$ .

$- (9 u^{2} + 16 u) - 1 \cdot 7 = 0$

Step 8.1.5

Factor $- 1$ out of $- (9 u^{2} + 16 u) - 1 (7)$ .

$- (9 u^{2} + 16 u + 7) = 0$

Step 8.2

Factor.

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Step 8.2.1

Factor by grouping.

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Step 8.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot 7 = 63$ and whose sum is $b = 16$ .

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Step 8.2.1.1.1

Factor $16$ out of $16 u$ .

$- (9 u^{2} + 16 u + 7) = 0$

Step 8.2.1.1.2

Rewrite $16$ as $7$ plus $9$

$- (9 u^{2} + (7 + 9) u + 7) = 0$

Step 8.2.1.1.3

Apply the distributive property.

$- (9 u^{2} + 7 u + 9 u + 7) = 0$

Step 8.2.1.2

Factor out the greatest common factor from each group.

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Step 8.2.1.2.1

Group the first two terms and the last two terms.

$- (9 u^{2} + 7 u + 9 u + 7) = 0$

Step 8.2.1.2.2

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

$- (u (9 u + 7) + 1 (9 u + 7)) = 0$

Step 8.2.1.3

Factor the polynomial by factoring out the greatest common factor, $9 u + 7$ .

$- ((9 u + 7) (u + 1)) = 0$

Step 8.2.2

Remove unnecessary parentheses.

$- (9 u + 7) (u + 1) = 0$

Step 9

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$9 u + 7 = 0$

$u + 1 = 0$

Step 10

Set $9 u + 7$ equal to $0$ and solve for $u$ .

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Step 10.1

Set $9 u + 7$ equal to $0$ .

$9 u + 7 = 0$

Step 10.2

Solve $9 u + 7 = 0$ for $u$ .

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Step 10.2.1

Subtract $7$ from both sides of the equation.

$9 u = - 7$

Step 10.2.2

Divide each term in $9 u = - 7$ by $9$ and simplify.

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Step 10.2.2.1

Divide each term in $9 u = - 7$ by $9$ .

$\frac{9 u}{9} = \frac{- 7}{9}$

Step 10.2.2.2

Simplify the left side.

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Step 10.2.2.2.1

Cancel the common factor of $9$ .

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Step 10.2.2.2.1.1

Cancel the common factor.

$\frac{9 u}{9} = \frac{- 7}{9}$

Step 10.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{- 7}{9}$

Step 10.2.2.3

Simplify the right side.

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Step 10.2.2.3.1

Move the negative in front of the fraction.

$u = - \frac{7}{9}$

Step 11

Set $u + 1$ equal to $0$ and solve for $u$ .

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Step 11.1

Set $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 11.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 12

The final solution is all the values that make $- (9 u + 7) (u + 1) = 0$ true.

$u = - \frac{7}{9}, - 1$

Step 13

Substitute $\sin (θ)$ for $u$ .

$\sin (θ) = - \frac{7}{9}, - 1$

Step 14

Set up each of the solutions to solve for $θ$ .

$\sin (θ) = - \frac{7}{9}$

$\sin (θ) = - 1$

Step 15

Solve for $θ$ in $\sin (θ) = - \frac{7}{9}$ .

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Step 15.1

Take the inverse sine of both sides of the equation to extract $θ$ from inside the sine.

$θ = \arcsin (- \frac{7}{9})$

Step 15.2

Simplify the right side.

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Step 15.2.1

Evaluate $\arcsin (- \frac{7}{9})$ .

$θ = - 51.05755873$

Step 15.3

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$θ = 360 + 51.05755873 + 180$

Step 15.4

Simplify the expression to find the second solution.

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Step 15.4.1

Subtract $360 °$ from $360 + 51.05755873 + 180 °$ .

$θ = 360 + 51.05755873 + 180 ° - 360 °$

Step 15.4.2

The resulting angle of $231.05755873 °$ is positive, less than $360 °$ , and coterminal with $360 + 51.05755873 + 180$ .

$θ = 231.05755873 °$

Step 15.5

Find the period of $\sin (θ)$ .

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Step 15.5.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

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

Step 15.5.2

Replace $b$ with $1$ in the formula for period.

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

Step 15.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Step 15.5.4

Divide $360$ by $1$ .

$360$

Step 15.6

Add $360$ to every negative angle to get positive angles.

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Step 15.6.1

Add $360$ to $- 51.05755873$ to find the positive angle.

$- 51.05755873 + 360$

Step 15.6.2

Subtract $51.05755873$ from $360$ .

$308.94244126$

Step 15.6.3

List the new angles.

$θ = 308.94244126$

Step 15.7

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 231.05755873 + 360 n, 308.94244126 + 360 n$ , for any integer $n$

Step 16

Solve for $θ$ in $\sin (θ) = - 1$ .

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Step 16.1

Take the inverse sine of both sides of the equation to extract $θ$ from inside the sine.

$θ = \arcsin (- 1)$

Step 16.2

Simplify the right side.

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Step 16.2.1

The exact value of $\arcsin (- 1)$ is $- 90$ .

$θ = - 90$

Step 16.3

$θ = 360 + 90 + 180$

Step 16.4

Simplify the expression to find the second solution.

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Step 16.4.1

Subtract $360 °$ from $360 + 90 + 180 °$ .

$θ = 360 + 90 + 180 ° - 360 °$

Step 16.4.2

The resulting angle of $270 °$ is positive, less than $360 °$ , and coterminal with $360 + 90 + 180$ .

$θ = 270 °$

Step 16.5

Find the period of $\sin (θ)$ .

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Step 16.5.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

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

Step 16.5.2

Replace $b$ with $1$ in the formula for period.

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

Step 16.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Step 16.5.4

Divide $360$ by $1$ .

$360$

Step 16.6

Add $360$ to every negative angle to get positive angles.

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Step 16.6.1

Add $360$ to $- 90$ to find the positive angle.

$- 90 + 360$

Step 16.6.2

Subtract $90$ from $360$ .

$270$

Step 16.6.3

List the new angles.

$θ = 270$

Step 16.7

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 270 + 360 n, 270 + 360 n$ , for any integer $n$

Step 17

List all of the solutions.

$θ = 231.05755873 + 360 n, 308.94244126 + 360 n, 270 + 360 n, 270 + 360 n$ , for any integer $n$

Step 18

Consolidate $270 + 360 n$ and $270 + 360 n$ to $270 + 360 n$ .

$θ = 231.05755873 + 360 n, 308.94244126 + 360 n, 270 + 360 n$ , for any integer $n$