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Popular >

1<= tan(x)<= 3^{1/2}

Solution

1 \leq tan (x) \leq 3^{\frac{1}{2}}

Solution

\frac{π}{4} + πn \leq x \leq \frac{π}{3} + πn

+2

Interval Notation

[\frac{π}{4} + πn, \frac{π}{3} + πn]

Decimal Notation

0.78539 \dots + πn \leq x \leq 1.04719 \dots + πn

Solution steps

1 \leq tan (x) \leq 3^{\frac{1}{2}}

If

a \leq u \leq b

then

a \leq u and u \leq b

1 \leq tan (x) and tan (x) \leq 3^{\frac{1}{2}}

1 \leq tan (x) : \frac{π}{4} + πn \leq x < \frac{π}{2} + πn

1 \leq tan (x)

Switch sides

tan (x) \geq 1

If

tan (x) \geq a

then

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

arctan (1) + πn \leq x < \frac{π}{2} + 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}{4} + πn \leq x < \frac{π}{2} + πn

tan (x) \leq 3^{\frac{1}{2}} : - \frac{π}{2} + πn < x \leq \frac{π}{3} + πn

tan (x) \leq 3^{\frac{1}{2}}

If

tan (x) \leq a

then

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

- \frac{π}{2} + πn < x \leq arctan (3^{\frac{1}{2}}) + πn

Simplify

arctan (3^{\frac{1}{2}}) : \frac{π}{3}

arctan (3^{\frac{1}{2}})

Use the following trivial identity:

arctan (3^{\frac{1}{2}}) = \frac{π}{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}{2} + πn < x \leq \frac{π}{3} + πn

Combine the intervals

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

Merge Overlapping Intervals

\frac{π}{4} + πn \leq x \leq \frac{π}{3} + πn

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