Precalculus Examples

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

Step 1

Replace $\sin^{2} (x)$ with $1 - \cos^{2} (x)$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Add $1$ and $5$ .

$6 - \cos^{2} (x) - 5 \cos (x) = 0$

Step 2.2

Factor by grouping.

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Step 2.2.1

Reorder terms.

$- \cos^{2} (x) - 5 \cos (x) + 6 = 0$

Step 2.2.2

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 $a \cdot c = - 1 \cdot 6 = - 6$ and whose sum is $b = - 5$ .

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Step 2.2.2.1

Factor $- 5$ out of $- 5 \cos (x)$ .

$- \cos^{2} (x) - 5 \cos (x) + 6 = 0$

Step 2.2.2.2

Rewrite $- 5$ as $1$ plus $- 6$

$- \cos^{2} (x) + (1 - 6) \cos (x) + 6 = 0$

Step 2.2.2.3

Apply the distributive property.

$- \cos^{2} (x) + 1 \cos (x) - 6 \cos (x) + 6 = 0$

Step 2.2.2.4

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

$- \cos^{2} (x) + \cos (x) - 6 \cos (x) + 6 = 0$

Step 2.2.3

Factor out the greatest common factor from each group.

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Step 2.2.3.1

Group the first two terms and the last two terms.

$- \cos^{2} (x) + \cos (x) - 6 \cos (x) + 6 = 0$

Step 2.2.3.2

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

$\cos (x) (- \cos (x) + 1) + 6 (- \cos (x) + 1) = 0$

Step 2.2.4

Factor the polynomial by factoring out the greatest common factor, $- \cos (x) + 1$ .

$(- \cos (x) + 1) (\cos (x) + 6) = 0$

Step 2.3

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

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

$\cos (x) + 6 = 0$

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$- \cos (x) = - 1$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.4.2.2.2.2

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

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

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$\cos (x) = 1$

Step 2.4.2.3

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

$x = \arccos (1)$

Step 2.4.2.4

Simplify the right side.

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Step 2.4.2.4.1

The exact value of $\arccos (1)$ is $0$ .

$x = 0$

Step 2.4.2.5

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.

$x = 2 π - 0$

Step 2.4.2.6

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 2.4.2.7

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

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Step 2.4.2.7.1

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

$\frac{2 π}{| b |}$

Step 2.4.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 2.4.2.7.3

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

$\frac{2 π}{1}$

Step 2.4.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.4.2.8

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

$x = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 2.5

Set $\cos (x) + 6$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $\cos (x) + 6$ equal to $0$ .

$\cos (x) + 6 = 0$

Step 2.5.2

Solve $\cos (x) + 6 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $6$ from both sides of the equation.

$\cos (x) = - 6$

Step 2.5.2.2

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 6$ does not fall in this range, there is no solution.

No solution

Step 2.6

The final solution is all the values that make $(- \cos (x) + 1) (\cos (x) + 6) = 0$ true.

$x = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 3

Consolidate the answers.

$x = 2 π n$ , for any integer $n$