Solve for x in Degrees tan(x)+ square root of 3=0

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Trigonometry

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

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

$\tan (x) = - \sqrt{3}$

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

$x = \arctan (- \sqrt{3})$

Simplify the right side.

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The exact value of $\arctan (- \sqrt{3})$ is $- 60$ .

$x = - 60$

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$x = - 60 - 180$

Simplify the expression to find the second solution.

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Add $360 °$ to $- 60 - 180 °$ .

$x = - 60 - 180 ° + 360 °$

The resulting angle of $120 °$ is positive and coterminal with $- 60 - 180$ .

$x = 120 °$

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

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The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

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

$\frac{180}{| 1 |}$

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

$\frac{180}{1}$

Divide $180$ by $1$ .

$180$

Add $180$ to every negative angle to get positive angles.

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Add $180$ to $- 60$ to find the positive angle.

$- 60 + 180$

Subtract $60$ from $180$ .

$120$

List the new angles.

$x = 120$

The period of the $\tan (x)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$x = 120 + 180 n, 120 + 180 n$ , for any integer $n$

Consolidate the answers.

$x = 120 + 180 n$ , for any integer $n$

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