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Calculus Examples
tan(θ)=-√3tan(θ)=−√3
Step 1
Take the inverse tangent of both sides of the equation to extract θ from inside the tangent.
θ=arctan(-√3)
Step 2
Step 2.1
The exact value of arctan(-√3) is -π3.
θ=-π3
θ=-π3
Step 3
The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from π to find the solution in the third quadrant.
θ=-π3-π
Step 4
Step 4.1
Add 2π to -π3-π.
θ=-π3-π+2π
Step 4.2
The resulting angle of 2π3 is positive and coterminal with -π3-π.
θ=2π3
θ=2π3
Step 5
Step 5.1
The period of the function can be calculated using π|b|.
π|b|
Step 5.2
Replace b with 1 in the formula for period.
π|1|
Step 5.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
π1
Step 5.4
Divide π by 1.
π
π
Step 6
Step 6.1
Add π to -π3 to find the positive angle.
-π3+π
Step 6.2
To write π as a fraction with a common denominator, multiply by 33.
π⋅33-π3
Step 6.3
Combine fractions.
Step 6.3.1
Combine π and 33.
π⋅33-π3
Step 6.3.2
Combine the numerators over the common denominator.
π⋅3-π3
π⋅3-π3
Step 6.4
Simplify the numerator.
Step 6.4.1
Move 3 to the left of π.
3⋅π-π3
Step 6.4.2
Subtract π from 3π.
2π3
2π3
Step 6.5
List the new angles.
θ=2π3
θ=2π3
Step 7
The period of the tan(θ) function is π so values will repeat every π radians in both directions.
θ=2π3+πn,2π3+πn, for any integer n
Step 8
Consolidate the answers.
θ=2π3+πn, for any integer n