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Popular >

tan(-3x)<1

Solution

tan (- 3 x) < 1

Solution

- \frac{π}{12} + \frac{π}{3} n < x < \frac{π}{6} + \frac{π}{3} n

Interval Notation

(- \frac{π}{12} + \frac{π}{3} n, \frac{π}{6} + \frac{π}{3} n)

Decimal Notation

- 0.26179 \dots + \frac{π}{3} n < x < 0.52359 \dots + \frac{π}{3} n

Solution steps

tan (- 3 x) < 1

Use the following identity:

tan (- x) = - tan (x)

- tan (3 x) < 1

Multiply both sides by

- 1

- tan (3 x) < 1

Multiply both sides by -1 (reverse the inequality)

(- tan (3 x)) (- 1) > 1 \cdot (- 1)

Simplify

tan (3 x) > - 1

tan (3 x) > - 1

tan (x) > a

then

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

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

a < u < b

then

a < u and u < b

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

arctan (- 1) + πn < 3 x : x > - \frac{π}{12} + \frac{πn}{3}

arctan (- 1) + πn < 3 x

Switch sides

3 x > arctan (- 1) + πn

Simplify

arctan (- 1) + πn : - \frac{π}{4} + πn

arctan (- 1) + πn

arctan (- 1) = - \frac{π}{4}

arctan (- 1)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 1) = - arctan (1)

= - arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

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}

= - \frac{π}{4} + πn

3 x > - \frac{π}{4} + πn

Divide both sides by

3

3 x > - \frac{π}{4} + πn

Divide both sides by

3

\frac{3 x}{3} > - \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} > - \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

- \frac{\frac{π}{4}}{3} + \frac{πn}{3} : - \frac{π}{12} + \frac{πn}{3}

- \frac{\frac{π}{4}}{3} + \frac{πn}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{π}{12}

= - \frac{π}{12} + \frac{πn}{3}

x > - \frac{π}{12} + \frac{πn}{3}

x > - \frac{π}{12} + \frac{πn}{3}

x > - \frac{π}{12} + \frac{πn}{3}

3 x < \frac{π}{2} + πn : x < \frac{π}{6} + \frac{πn}{3}

3 x < \frac{π}{2} + πn

Divide both sides by

3

3 x < \frac{π}{2} + πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{2}}{3} + \frac{πn}{3} : \frac{π}{6} + \frac{πn}{3}

\frac{\frac{π}{2}}{3} + \frac{πn}{3}

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{πn}{3}

x < \frac{π}{6} + \frac{πn}{3}

x < \frac{π}{6} + \frac{πn}{3}

x < \frac{π}{6} + \frac{πn}{3}

Combine the intervals

x > - \frac{π}{12} + \frac{πn}{3} and x < \frac{π}{6} + \frac{πn}{3}

Merge Overlapping Intervals

- \frac{π}{12} + \frac{π}{3} n < x < \frac{π}{6} + \frac{π}{3} n

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