Trigonometry Examples

$16 \sin (x) = 4 \cos^{2} (x) - 11$

Step 1

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

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Step 1.1

Subtract $4 \cos^{2} (x)$ from both sides of the equation.

$16 \sin (x) - 4 \cos^{2} (x) = - 11$

Step 1.2

Add $11$ to both sides of the equation.

$16 \sin (x) - 4 \cos^{2} (x) + 11 = 0$

Step 2

Replace the $- 4 \cos^{2} (x)$ with $- 4 (1 - \sin^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$16 \sin (x) - 4 (1 - \sin^{2} (x)) + 11 = 0$

Step 3

Simplify each term.

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Step 3.1

Apply the distributive property.

$16 \sin (x) - 4 \cdot 1 - 4 (- \sin^{2} (x)) + 11 = 0$

Step 3.2

Multiply $- 4$ by $1$ .

$16 \sin (x) - 4 - 4 (- \sin^{2} (x)) + 11 = 0$

Step 3.3

Multiply $- 1$ by $- 4$ .

$16 \sin (x) - 4 + 4 \sin^{2} (x) + 11 = 0$

Step 4

Add $- 4$ and $11$ .

$16 \sin (x) + 4 \sin^{2} (x) + 7 = 0$

Step 5

Reorder the polynomial.

$4 \sin^{2} (x) + 16 \sin (x) + 7 = 0$

Step 6

Substitute $u$ for $\sin (x)$ .

$4 {(u)}^{2} + 16 (u) + 7 = 0$

Step 7

Factor by grouping.

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Step 7.1

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 $a \cdot c = 4 \cdot 7 = 28$ and whose sum is $b = 16$ .

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Step 7.1.1

Factor $16$ out of $16 u$ .

$4 u^{2} + 16 u + 7 = 0$

Step 7.1.2

Rewrite $16$ as $2$ plus $14$

$4 u^{2} + (2 + 14) u + 7 = 0$

Step 7.1.3

Apply the distributive property.

$4 u^{2} + 2 u + 14 u + 7 = 0$

Step 7.2

Factor out the greatest common factor from each group.

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Step 7.2.1

Group the first two terms and the last two terms.

$4 u^{2} + 2 u + 14 u + 7 = 0$

Step 7.2.2

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

$2 u (2 u + 1) + 7 (2 u + 1) = 0$

Step 7.3

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

$(2 u + 1) (2 u + 7) = 0$

Step 8

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

$2 u + 1 = 0$

$2 u + 7 = 0$

Step 9

Set $2 u + 1$ equal to $0$ and solve for $u$ .

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Step 9.1

Set $2 u + 1$ equal to $0$ .

$2 u + 1 = 0$

Step 9.2

Solve $2 u + 1 = 0$ for $u$ .

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Step 9.2.1

Subtract $1$ from both sides of the equation.

$2 u = - 1$

Step 9.2.2

Divide each term in $2 u = - 1$ by $2$ and simplify.

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Step 9.2.2.1

Divide each term in $2 u = - 1$ by $2$ .

$\frac{2 u}{2} = \frac{- 1}{2}$

Step 9.2.2.2

Simplify the left side.

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Step 9.2.2.2.1

Cancel the common factor of $2$ .

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Step 9.2.2.2.1.1

Cancel the common factor.

$\frac{2 u}{2} = \frac{- 1}{2}$

Step 9.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{- 1}{2}$

Step 9.2.2.3

Simplify the right side.

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Step 9.2.2.3.1

Move the negative in front of the fraction.

$u = - \frac{1}{2}$

Step 10

Set $2 u + 7$ equal to $0$ and solve for $u$ .

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Step 10.1

Set $2 u + 7$ equal to $0$ .

$2 u + 7 = 0$

Step 10.2

Solve $2 u + 7 = 0$ for $u$ .

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Step 10.2.1

Subtract $7$ from both sides of the equation.

$2 u = - 7$

Step 10.2.2

Divide each term in $2 u = - 7$ by $2$ and simplify.

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Step 10.2.2.1

Divide each term in $2 u = - 7$ by $2$ .

$\frac{2 u}{2} = \frac{- 7}{2}$

Step 10.2.2.2

Simplify the left side.

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Step 10.2.2.2.1

Cancel the common factor of $2$ .

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Step 10.2.2.2.1.1

Cancel the common factor.

$\frac{2 u}{2} = \frac{- 7}{2}$

Step 10.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{- 7}{2}$

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.

$u = - \frac{7}{2}$

Step 11

The final solution is all the values that make $(2 u + 1) (2 u + 7) = 0$ true.

$u = - \frac{1}{2}, - \frac{7}{2}$

Step 12

Substitute $\sin (x)$ for $u$ .

$\sin (x) = - \frac{1}{2}, - \frac{7}{2}$

Step 13

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

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

$\sin (x) = - \frac{7}{2}$

Step 14

Solve for $x$ in $\sin (x) = - \frac{1}{2}$ .

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Step 14.1

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

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

Step 14.2

Simplify the right side.

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Step 14.2.1

The exact value of $\arcsin (- \frac{1}{2})$ is $- \frac{π}{6}$ .

$x = - \frac{π}{6}$

Step 14.3

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 14.4

Simplify the expression to find the second solution.

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Step 14.4.1

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$x = 2 π + \frac{π}{6} + π - 2 π$

Step 14.4.2

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 14.5

Find the period of $\sin (x)$ .

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Step 14.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 14.5.2

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

$\frac{2 π}{| 1 |}$

Step 14.5.3

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

$\frac{2 π}{1}$

Step 14.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 14.6

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

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Step 14.6.1

Add $2 π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

Step 14.6.2

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 14.6.3

Combine fractions.

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Step 14.6.3.1

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 14.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 14.6.4

Simplify the numerator.

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Step 14.6.4.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Step 14.6.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

Step 14.6.5

List the new angles.

$x = \frac{11 π}{6}$

Step 14.7

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 $n$

Step 15

Solve for $x$ in $\sin (x) = - \frac{7}{2}$ .

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Step 15.1

The range of sine is $- 1 \leq y \leq 1$ . Since $- \frac{7}{2}$ does not fall in this range, there is no solution.

No solution

Step 16

List all of the solutions.

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