Precalculus Examples

$10 \sin^{2} (x) = 10 + 5 \cos (x)$

Step 1

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

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Step 1.1

Subtract $10$ from both sides of the equation.

$10 \sin^{2} (x) - 10 = 5 \cos (x)$

Step 1.2

Subtract $5 \cos (x)$ from both sides of the equation.

$10 \sin^{2} (x) - 10 - 5 \cos (x) = 0$

Step 2

Simplify $10 \sin^{2} (x) - 10 - 5 \cos (x)$ .

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Step 2.1

Reorder $10 \sin^{2} (x)$ and $- 10$ .

$- 10 + 10 \sin^{2} (x) - 5 \cos (x) = 0$

Step 2.2

Factor $- 10$ out of $- 10$ .

$- 10 (1) + 10 \sin^{2} (x) - 5 \cos (x) = 0$

Step 2.3

Factor $- 10$ out of $10 \sin^{2} (x)$ .

$- 10 (1) - 10 (- \sin^{2} (x)) - 5 \cos (x) = 0$

Step 2.4

Factor $- 10$ out of $- 10 (1) - 10 (- \sin^{2} (x))$ .

$- 10 (1 - \sin^{2} (x)) - 5 \cos (x) = 0$

Step 2.5

Apply pythagorean identity.

$- 10 \cos^{2} (x) - 5 \cos (x) = 0$

Step 3

Solve for $x$ .

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Step 3.1

Factor the left side of the equation.

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Step 3.1.1

Let $u = \cos (x)$ . Substitute $u$ for all occurrences of $\cos (x)$ .

$- 10 u^{2} - 5 u = 0$

Step 3.1.2

Factor $5 u$ out of $- 10 u^{2} - 5 u$ .

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Step 3.1.2.1

Factor $5 u$ out of $- 10 u^{2}$ .

$5 u (- 2 u) - 5 u = 0$

Step 3.1.2.2

Factor $5 u$ out of $- 5 u$ .

$5 u (- 2 u) + 5 u (- 1) = 0$

Step 3.1.2.3

Factor $5 u$ out of $5 u (- 2 u) + 5 u (- 1)$ .

$5 u (- 2 u - 1) = 0$

Step 3.1.3

Replace all occurrences of $u$ with $\cos (x)$ .

$5 \cos (x) (- 2 \cos (x) - 1) = 0$

Step 3.2

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

$\cos (x) = 0$

$- 2 \cos (x) - 1 = 0$

Step 3.3

Set $\cos (x)$ equal to $0$ and solve for $x$ .

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Step 3.3.1

Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Step 3.3.2

Solve $\cos (x) = 0$ for $x$ .

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Step 3.3.2.1

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

$x = \arccos (0)$

Step 3.3.2.2

Simplify the right side.

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Step 3.3.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 3.3.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{2}$

Step 3.3.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 3.3.2.4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.3.2.4.2

Combine fractions.

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Step 3.3.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 3.3.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 3.3.2.4.3

Simplify the numerator.

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Step 3.3.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 3.3.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 3.3.2.5

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

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Step 3.3.2.5.1

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

$\frac{2 π}{| b |}$

Step 3.3.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 3.3.2.5.3

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

$\frac{2 π}{1}$

Step 3.3.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.3.2.6

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Step 3.4

Set $- 2 \cos (x) - 1$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $- 2 \cos (x) - 1$ equal to $0$ .

$- 2 \cos (x) - 1 = 0$

Step 3.4.2

Solve $- 2 \cos (x) - 1 = 0$ for $x$ .

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Step 3.4.2.1

Add $1$ to both sides of the equation.

$- 2 \cos (x) = 1$

Step 3.4.2.2

Divide each term in $- 2 \cos (x) = 1$ by $- 2$ and simplify.

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Step 3.4.2.2.1

Divide each term in $- 2 \cos (x) = 1$ by $- 2$ .

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

Step 3.4.2.2.2

Simplify the left side.

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Step 3.4.2.2.2.1

Cancel the common factor of $- 2$ .

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Step 3.4.2.2.2.1.1

Cancel the common factor.

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

Step 3.4.2.2.2.1.2

Divide $\cos (x)$ by $1$ .

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

Step 3.4.2.2.3

Simplify the right side.

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Step 3.4.2.2.3.1

Move the negative in front of the fraction.

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

Step 3.4.2.3

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

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

Step 3.4.2.4

Simplify the right side.

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Step 3.4.2.4.1

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$x = \frac{2 π}{3}$

Step 3.4.2.5

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - \frac{2 π}{3}$

Step 3.4.2.6

Simplify $2 π - \frac{2 π}{3}$ .

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Step 3.4.2.6.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Step 3.4.2.6.2

Combine fractions.

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Step 3.4.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Step 3.4.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 3.4.2.6.3

Simplify the numerator.

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Step 3.4.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 3.4.2.6.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 3.4.2.7

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

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Step 3.4.2.7.1

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

$\frac{2 π}{| b |}$

Step 3.4.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 3.4.2.7.3

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

$\frac{2 π}{1}$

Step 3.4.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.4.2.8

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Step 3.5

The final solution is all the values that make $5 \cos (x) (- 2 \cos (x) - 1) = 0$ true.

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Step 4

Consolidate $\frac{π}{2} + 2 π n$ and $\frac{3 π}{2} + 2 π n$ to $\frac{π}{2} + π n$ .

$x = \frac{π}{2} + π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$