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Popular >

tan^2(x)<= 3

Solution

tan^{2} (x) \leq 3

Solution

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

+2

Interval Notation

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

Decimal Notation

πn \leq x \leq 1.04719 \dots + πn or 2.09439 \dots + πn \leq x < 3.14159 \dots + πn

Solution steps

tan^{2} (x) \leq 3

For

u^{n} \leq a

, if

n

is even then

- 3 \leq tan (x) \leq 3

If

a \leq u \leq b

then

a \leq u and u \leq b

- 3 \leq tan (x) and tan (x) \leq 3

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

- 3 \leq tan (x)

Switch sides

tan (x) \geq - 3

If

tan (x) \geq a

then

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

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

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

tan (x) \leq 3

If

tan (x) \leq a

then

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

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

Simplify

arctan (3) : \frac{π}{3}

arctan (3)

Use the following trivial identity:

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

Merge Overlapping Intervals

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

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