Precalculus Examples

Solve for ? tan(theta)=( square root of 3)/3
tan(θ)=33tan(θ)=33
Step 1
Take the inverse tangent of both sides of the equation to extract θθ from inside the tangent.
θ=arctan(33)θ=arctan(33)
Step 2
Simplify the right side.
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Step 2.1
The exact value of arctan(33)arctan(33) is π6π6.
θ=π6θ=π6
θ=π6θ=π6
Step 3
The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from ππ to find the solution in the fourth quadrant.
θ=π+π6θ=π+π6
Step 4
Simplify π+π6π+π6.
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Step 4.1
To write ππ as a fraction with a common denominator, multiply by 6666.
θ=π66+π6θ=π66+π6
Step 4.2
Combine fractions.
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Step 4.2.1
Combine ππ and 6666.
θ=π66+π6θ=π66+π6
Step 4.2.2
Combine the numerators over the common denominator.
θ=π6+π6θ=π6+π6
θ=π6+π6θ=π6+π6
Step 4.3
Simplify the numerator.
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Step 4.3.1
Move 66 to the left of ππ.
θ=6π+π6θ=6π+π6
Step 4.3.2
Add 6π6π and ππ.
θ=7π6θ=7π6
θ=7π6θ=7π6
θ=7π6θ=7π6
Step 5
Find the period of tan(θ).
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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
The period of the tan(θ) function is π so values will repeat every π radians in both directions.
θ=π6+πn,7π6+πn, for any integer n
Step 7
Consolidate the answers.
θ=π6+πn, for any integer n
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