Solve for x cos(x)^2-cos(x)=2

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

Step 1

Subtract $2$ from both sides of the equation.

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

Step 2

Factor $\cos^{2} (x) - \cos (x) - 2$ using the AC method.

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 $- 1$ .

$- 2, 1$

Write the factored form using these integers.

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

Step 3

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

$\cos (x) - 2 = 0$

$\cos (x) + 1 = 0$

Step 4

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

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

$\cos (x) - 2 = 0$

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

Add $2$ to both sides of the equation.

$\cos (x) = 2$

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

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

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

$\cos (x) + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

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

$x = \arccos (- 1)$

Simplify the right side.

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

$x = π$

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.

$x = 2 π - π$

Subtract $π$ from $2 π$ .

$x = π$

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

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

$x = π + 2 π n$ , for any integer $n$

Step 6

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

$x = π + 2 π n$ , for any integer $n$

Solve for x cos(x)^2-cos(x)=2

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