Precalculus Examples

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

Step 1

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

${(u)}^{2} - (u) - 6 = 0$

Step 2

Factor $u^{2} - u - 6$ using the AC method.

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Step 2.1

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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 2.2

Write the factored form using these integers.

$(u - 3) (u + 2) = 0$

Step 3

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

$u - 3 = 0$

$u + 2 = 0$

Step 4

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

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Step 4.1

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

$u - 3 = 0$

Step 4.2

Add $3$ to both sides of the equation.

$u = 3$

Step 5

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

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Step 5.1

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

$u + 2 = 0$

Step 5.2

Subtract $2$ from both sides of the equation.

$u = - 2$

Step 6

The final solution is all the values that make $(u - 3) (u + 2) = 0$ true.

$u = 3, - 2$

Step 7

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

$\cos (x) = 3, - 2$

Step 8

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

$\cos (x) = 3$

$\cos (x) = - 2$

Step 9

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

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Step 9.1

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

No solution

Step 10

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

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Step 10.1

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 2$ does not fall in this range, there is no solution.

No solution