Solve for θ tan(theta)=4/3

$\tan (θ) = \frac{4}{3}$

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

$θ = \arctan (\frac{4}{3})$

Simplify the right side.

Evaluate $\arctan (\frac{4}{3})$ .

$θ = 0.92729521$

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

$θ = (3.14159265) + 0.92729521$

Solve for $θ$ .

Remove parentheses.

$θ = 3.14159265 + 0.92729521$

Remove parentheses.

$θ = (3.14159265) + 0.92729521$

Add $3.14159265$ and $0.92729521$ .

$θ = 4.06888787$

Find the period of $\tan (θ)$ .

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

$θ = 0.92729521 + π n, 4.06888787 + π n$ , for any integer $n$

Consolidate $0.92729521 + π n$ and $4.06888787 + π n$ to $0.92729521 + π n$ .

$θ = 0.92729521 + π n$ , for any integer $n$

Solve for θ tan(theta)=4/3

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