Trigonometry Examples

$4 \cos^{2} (x) - 4 \sin (x) - 5 = 0$

Step 1

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

$4 (1 - \sin^{2} (x)) - 4 \sin (x) - 5 = 0$

Step 2

Simplify each term.

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Step 2.1

Apply the distributive property.

$4 \cdot 1 + 4 (- \sin^{2} (x)) - 4 \sin (x) - 5 = 0$

Step 2.2

Multiply $4$ by $1$ .

$4 + 4 (- \sin^{2} (x)) - 4 \sin (x) - 5 = 0$

Step 2.3

Multiply $- 1$ by $4$ .

$4 - 4 \sin^{2} (x) - 4 \sin (x) - 5 = 0$

Step 3

Subtract $5$ from $4$ .

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

Step 4

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

$- 4 {(u)}^{2} - 4 u - 1 = 0$

Step 5

Factor the left side of the equation.

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Step 5.1

Factor $- 1$ out of $- 4 u^{2} - 4 u - 1$ .

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Step 5.1.1

Factor $- 1$ out of $- 4 u^{2}$ .

$- (4 u^{2}) - 4 u - 1 = 0$

Step 5.1.2

Factor $- 1$ out of $- 4 u$ .

$- (4 u^{2}) - (4 u) - 1 = 0$

Step 5.1.3

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

$- (4 u^{2}) - (4 u) - 1 \cdot 1 = 0$

Step 5.1.4

Factor $- 1$ out of $- (4 u^{2}) - (4 u)$ .

$- (4 u^{2} + 4 u) - 1 \cdot 1 = 0$

Step 5.1.5

Factor $- 1$ out of $- (4 u^{2} + 4 u) - 1 (1)$ .

$- (4 u^{2} + 4 u + 1) = 0$

Step 5.2

Factor using the perfect square rule.

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Step 5.2.1

Rewrite $4 u^{2}$ as ${(2 u)}^{2}$ .

$- ({(2 u)}^{2} + 4 u + 1) = 0$

Step 5.2.2

Rewrite $1$ as $1^{2}$ .

$- ({(2 u)}^{2} + 4 u + 1^{2}) = 0$

Step 5.2.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 u = 2 \cdot (2 u) \cdot 1$

Step 5.2.4

Rewrite the polynomial.

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

Step 5.2.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 u$ and $b = 1$ .

$- {(2 u + 1)}^{2} = 0$

Step 6

Divide each term in $- {(2 u + 1)}^{2} = 0$ by $- 1$ and simplify.

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Step 6.1

Divide each term in $- {(2 u + 1)}^{2} = 0$ by $- 1$ .

$\frac{- {(2 u + 1)}^{2}}{- 1} = \frac{0}{- 1}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Dividing two negative values results in a positive value.

$\frac{{(2 u + 1)}^{2}}{1} = \frac{0}{- 1}$

Step 6.2.2

Divide ${(2 u + 1)}^{2}$ by $1$ .

${(2 u + 1)}^{2} = \frac{0}{- 1}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Divide $0$ by $- 1$ .

${(2 u + 1)}^{2} = 0$

Step 7

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

$2 u + 1 = 0$

Step 8

Solve for $u$ .

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Step 8.1

Subtract $1$ from both sides of the equation.

$2 u = - 1$

Step 8.2

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

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Step 8.2.1

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

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

Step 8.2.2

Simplify the left side.

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Step 8.2.2.1

Cancel the common factor of $2$ .

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Step 8.2.2.1.1

Cancel the common factor.

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

Step 8.2.2.1.2

Divide $u$ by $1$ .

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

Step 8.2.3

Simplify the right side.

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Step 8.2.3.1

Move the negative in front of the fraction.

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

Step 9

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

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

Step 10

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

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

Step 11

Simplify the right side.

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Step 11.1

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

$x = - \frac{π}{6}$

Step 12

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 13

Simplify the expression to find the second solution.

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Step 13.1

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

$x = 2 π + \frac{π}{6} + π - 2 π$

Step 13.2

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

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

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Step 14.1

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

$\frac{2 π}{| b |}$

Step 14.2

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

$\frac{2 π}{| 1 |}$

Step 14.3

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

$\frac{2 π}{1}$

Step 14.4

Divide $2 π$ by $1$ .

$2 π$

Step 15

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

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Step 15.1

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

$- \frac{π}{6} + 2 π$

Step 15.2

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 15.3

Combine fractions.

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Step 15.3.1

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 15.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 15.4

Simplify the numerator.

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Step 15.4.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Step 15.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

Step 15.5

List the new angles.

$x = \frac{11 π}{6}$

Step 16

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 $n$