Precalculus Examples

$\cos^{2} (x) - \cos (x) = 12$

Step 1

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

${(u)}^{2} - (u) = 12$

Step 2

Subtract $12$ from both sides of the equation.

$u^{2} - u - 12 = 0$

Step 3

Factor $u^{2} - u - 12$ using the AC method.

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Step 3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 3.2

Write the factored form using these integers.

$(u - 4) (u + 3) = 0$

Step 4

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

$u - 4 = 0$

$u + 3 = 0$

Step 5

Set $u - 4$ equal to $0$ and solve for $u$ .

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Step 5.1

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

Step 5.2

Add $4$ to both sides of the equation.

$u = 4$

Step 6

Set $u + 3$ equal to $0$ and solve for $u$ .

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Step 6.1

Set $u + 3$ equal to $0$ .

$u + 3 = 0$

Step 6.2

Subtract $3$ from both sides of the equation.

$u = - 3$

Step 7

The final solution is all the values that make $(u - 4) (u + 3) = 0$ true.

$u = 4, - 3$

Step 8

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

$\cos (x) = 4, - 3$

Step 9

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

$\cos (x) = 4$

$\cos (x) = - 3$

Step 10

Solve for $x$ in $\cos (x) = 4$ .

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Step 10.1

The range of cosine is $- 1 \leq y \leq 1$ . Since $4$ does not fall in this range, there is no solution.

No solution

Step 11

Solve for $x$ in $\cos (x) = - 3$ .

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Step 11.1

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 3$ does not fall in this range, there is no solution.

No solution