Solve for x in Degrees cos(x)+ square root of 3=-cos(x)

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Trigonometry

$\cos (x) + \sqrt{3} = - \cos (x)$

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

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Add $\cos (x)$ to both sides of the equation.

$\cos (x) + \sqrt{3} + \cos (x) = 0$

Add $\cos (x)$ and $\cos (x)$ .

$2 \cos (x) + \sqrt{3} = 0$

Subtract $\sqrt{3}$ from both sides of the equation.

$2 \cos (x) = - \sqrt{3}$

Divide each term in $2 \cos (x) = - \sqrt{3}$ by $2$ and simplify.

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Divide each term in $2 \cos (x) = - \sqrt{3}$ by $2$ .

$\frac{2 \cos (x)}{2} = \frac{- \sqrt{3}}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \cos (x)}{2} = \frac{- \sqrt{3}}{2}$

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

$\cos (x) = \frac{- \sqrt{3}}{2}$

Simplify the right side.

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Move the negative in front of the fraction.

$\cos (x) = - \frac{\sqrt{3}}{2}$

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

$x = \arccos (- \frac{\sqrt{3}}{2})$

Simplify the right side.

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The exact value of $\arccos (- \frac{\sqrt{3}}{2})$ is $150$ .

$x = 150$

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the third quadrant.

$x = 360 - 150$

Subtract $150$ from $360$ .

$x = 210$

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

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

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

$\frac{360}{| 1 |}$

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

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\cos (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 150 + 360 n, 210 + 360 n$ , for any integer $n$

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