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Popular >

-sqrt(3)<tan(θ)<1

Solution

- 3 < tan (θ) < 1

Solution

πn \leq θ < \frac{π}{4} + πn or \frac{2 π}{3} + πn < θ < π + πn

+2

Interval Notation

[πn, \frac{π}{4} + πn) \cup (\frac{2 π}{3} + πn, π + πn)

Decimal Notation

πn \leq θ < 0.78539 \dots + πn or 2.09439 \dots + πn < θ < 3.14159 \dots + πn

Solution steps

- 3 < tan (θ) < 1

If

a < u < b

then

a < u and u < b

- 3 < tan (θ) and tan (θ) < 1

- 3 < tan (θ) : - \frac{π}{3} + πn < θ < \frac{π}{2} + πn

- 3 < tan (θ)

Switch sides

tan (θ) > - 3

If

tan (x) > a

then

arctan (a) + πn < x < \frac{π}{2} + πn

arctan (- 3) + πn < θ < \frac{π}{2} + πn

Simplify

arctan (- 3) : - \frac{π}{3}

arctan (- 3)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 3) = - arctan (3)

= - arctan (3)

Use the following trivial identity:

arctan (3) = \frac{π}{3}

arctan (3)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{3}

= - \frac{π}{3}

- \frac{π}{3} + πn < θ < \frac{π}{2} + πn

tan (θ) < 1 : - \frac{π}{2} + πn < θ < \frac{π}{4} + πn

tan (θ) < 1

If

tan (x) < a

then

- \frac{π}{2} + πn < x < arctan (a) + πn

- \frac{π}{2} + πn < θ < arctan (1) + πn

Simplify

arctan (1) : \frac{π}{4}

arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

- \frac{π}{2} + πn < θ < \frac{π}{4} + πn

Combine the intervals

- \frac{π}{3} + πn < θ < \frac{π}{2} + πn and - \frac{π}{2} + πn < θ < \frac{π}{4} + πn

Merge Overlapping Intervals

πn \leq θ < \frac{π}{4} + πn or \frac{2 π}{3} + πn < θ < π + πn

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