Solve for x cos(x)+2=3cos(x)

$\cos (x) + 2 = 3 \cos (x)$

Move all terms containing $\cos (x)$ to the left side of the equation.

Subtract $3 \cos (x)$ from both sides of the equation.

$\cos (x) + 2 - 3 \cos (x) = 0$

Subtract $3 \cos (x)$ from $\cos (x)$ .

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

Subtract $2$ from both sides of the equation.

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

Divide each term in $- 2 \cos (x) = - 2$ by $- 2$ and simplify.

Divide each term in $- 2 \cos (x) = - 2$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Divide $- 2$ by $- 2$ .

$\cos (x) = 1$

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 $0$ .

$x = 0$

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.

$x = 2 π - 0$

Subtract $0$ from $2 π$ .

$x = 2 π$

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.

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

Consolidate the answers.

$x = 2 π n$ , for any integer $n$

Solve for x cos(x)+2=3cos(x)

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