Solve for c in Degrees -3tan(c)+3=tan(c)-1

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Trigonometry

$- 3 \tan (c) + 3 = \tan (c) - 1$

Move all terms containing $\tan (c)$ to the left side of the equation.

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Subtract $\tan (c)$ from both sides of the equation.

$- 3 \tan (c) + 3 - \tan (c) = - 1$

Subtract $\tan (c)$ from $- 3 \tan (c)$ .

$- 4 \tan (c) + 3 = - 1$

Move all terms not containing $\tan (c)$ to the right side of the equation.

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Subtract $3$ from both sides of the equation.

$- 4 \tan (c) = - 1 - 3$

Subtract $3$ from $- 1$ .

$- 4 \tan (c) = - 4$

Divide each term in $- 4 \tan (c) = - 4$ by $- 4$ and simplify.

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Divide each term in $- 4 \tan (c) = - 4$ by $- 4$ .

$\frac{- 4 \tan (c)}{- 4} = \frac{- 4}{- 4}$

Simplify the left side.

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Cancel the common factor of $- 4$ .

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Cancel the common factor.

$\frac{- 4 \tan (c)}{- 4} = \frac{- 4}{- 4}$

Divide $\tan (c)$ by $1$ .

$\tan (c) = \frac{- 4}{- 4}$

Simplify the right side.

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Divide $- 4$ by $- 4$ .

$\tan (c) = 1$

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

$c = \arctan (1)$

Simplify the right side.

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The exact value of $\arctan (1)$ is $45$ .

$c = 45$

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

$c = 180 + 45$

Add $180$ and $45$ .

$c = 225$

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

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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 $1$ .

$180$

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

$c = 45 + 180 n, 225 + 180 n$ , for any integer $n$

Consolidate the answers.

$c = 45 + 180 n$ , for any integer $n$

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