Trigonometry Examples

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

Step 1

Factor the left side of the equation.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 1.2.2

Write the factored form using these integers.

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

Step 1.3

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

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

Step 2

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

$\cos (x) - 1 = 0$

$\cos (x) + 2 = 0$

Step 3

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

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Step 3.1

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

$\cos (x) - 1 = 0$

Step 3.2

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

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Step 3.2.1

Add $1$ to both sides of the equation.

$\cos (x) = 1$

Step 3.2.2

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

$x = \arccos (1)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

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

$x = 0$

Step 3.2.4

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 3.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 3.2.6

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

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Step 3.2.6.1

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

$\frac{2 π}{| b |}$

Step 3.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

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

$\frac{2 π}{1}$

Step 3.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.7

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 4

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

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Step 4.1

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

$\cos (x) + 2 = 0$

Step 4.2

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

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Step 4.2.1

Subtract $2$ from both sides of the equation.

$\cos (x) = - 2$

Step 4.2.2

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

No solution

Step 5

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

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

Step 6

Consolidate the answers.

$x = 2 π n$ , for any integer $n$