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Popular >

2pi>sqrt(3)tan(θ)+1>= 0

Solution

2 π > 3 tan (θ) + 1 \geq 0

Solution

πn \leq θ < arctan (\frac{3 ( 2 π - 1 )}{3}) + πn or \frac{5 π}{6} + πn \leq θ < π + πn

Interval Notation

[πn, arctan (\frac{3 ( 2 π - 1 )}{3}) + πn) \cup [\frac{5 π}{6} + πn, π + πn)

Decimal Notation

πn \leq θ < 1.25399 \dots + πn or 2.61799 \dots + πn \leq θ < 3.14159 \dots + πn

Solution steps

2 π > 3 tan (θ) + 1 \geq 0

a > u \geq b

then

a > u and u \geq b

2 π > 3 tan (θ) + 1 and 3 tan (θ) + 1 \geq 0

2 π > 3 tan (θ) + 1 : - \frac{π}{2} + πn < θ < arctan (\frac{3 ( 2 π - 1 )}{3}) + πn

2 π > 3 tan (θ) + 1

Switch sides

3 tan (θ) + 1 < 2 π

Move

1

to the right side

3 tan (θ) + 1 < 2 π

Subtract

1

from both sides

3 tan (θ) + 1 - 1 < 2 π - 1

Simplify

3 tan (θ) < 2 π - 1

3 tan (θ) < 2 π - 1

Divide both sides by

3

3 tan (θ) < 2 π - 1

Divide both sides by

3

\frac{3 tan ( θ )}{3} < \frac{2 π}{3} - \frac{1}{3}

Simplify

\frac{3 tan ( θ )}{3} < \frac{2 π}{3} - \frac{1}{3}

Simplify

\frac{3 tan ( θ )}{3} : tan (θ)

\frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

\frac{2 π}{3} - \frac{1}{3} : \frac{3 ( 2 π - 1 )}{3}

\frac{2 π}{3} - \frac{1}{3}

Apply rule

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

= \frac{2 π - 1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{( 2 π - 1 ) 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3 ( 2 π - 1 )}{3}

tan (θ) < \frac{3 ( 2 π - 1 )}{3}

tan (θ) < \frac{3 ( 2 π - 1 )}{3}

tan (θ) < \frac{3 ( 2 π - 1 )}{3}

tan (x) < a

then

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

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

3 tan (θ) + 1 \geq 0 : - \frac{π}{6} + πn \leq θ < \frac{π}{2} + πn

3 tan (θ) + 1 \geq 0

Move

1

to the right side

3 tan (θ) + 1 \geq 0

Subtract

1

from both sides

3 tan (θ) + 1 - 1 \geq 0 - 1

Simplify

3 tan (θ) \geq - 1

3 tan (θ) \geq - 1

Divide both sides by

3

3 tan (θ) \geq - 1

Divide both sides by

3

\frac{3 tan ( θ )}{3} \geq \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} \geq \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} : tan (θ)

\frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

tan (θ) \geq - \frac{3}{3}

tan (θ) \geq - \frac{3}{3}

tan (θ) \geq - \frac{3}{3}

tan (x) \geq a

then

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

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

Simplify

arctan (- \frac{3}{3}) : - \frac{π}{6}

arctan (- \frac{3}{3})

Use the following property:

arctan (- x) = - arctan (x)

arctan (- \frac{3}{3}) = - arctan (\frac{3}{3})

= - arctan (\frac{3}{3})

Use the following trivial identity:

arctan (\frac{3}{3}) = \frac{π}{6}

arctan (\frac{3}{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}{6}

= - \frac{π}{6}

- \frac{π}{6} + πn \leq θ < \frac{π}{2} + πn

Combine the intervals

- \frac{π}{2} + πn < θ < arctan (\frac{3 ( 2 π - 1 )}{3}) + πn and - \frac{π}{6} + πn \leq θ < \frac{π}{2} + πn

Merge Overlapping Intervals

πn \leq θ < arctan (\frac{3 ( 2 π - 1 )}{3}) + πn or \frac{5 π}{6} + πn \leq θ < π + πn

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