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Popular Trigonometry >

3tan^2(x)+sqrt(3)tan(x)<= 0

Solution

3 tan^{2} (x) + 3 tan (x) \leq 0

Solution

\frac{5 π}{6} + πn \leq x < π + πn

Interval Notation

[\frac{5 π}{6} + πn, π + πn)

Decimal

2.61799 \dots + πn \leq x < 3.14159 \dots + πn

Solution steps

3 tan^{2} (x) + 3 tan (x) \leq 0

Let:

u = tan (x)

3 u^{2} + 3 u \leq 0

3 u^{2} + 3 u \leq 0 : - \frac{3}{3} \leq u \leq 0

3 u^{2} + 3 u \leq 0

Factor

3 u^{2} + 3 u : u (3 u + 3)

3 u^{2} + 3 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 3 uu + 3 u

Factor out common term

u

= u (3 u + 1 \cdot 3)

Multiply the numbers:

1 \cdot 3 = 3

= u (3 u + 3)

u (3 u + 3) \leq 0

Identify the intervals

Find the signs of the factors of

u (3 u + 3)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

3 u + 3

3 u + 3 = 0 : u = - \frac{3}{3}

3 u + 3 = 0

Move

3

to the right side

3 u + 3 = 0

Subtract

3

from both sides

3 u + 3 - 3 = 0 - 3

Simplify

3 u = - 3

3 u = - 3

Divide both sides by

3

3 u = - 3

Divide both sides by

3

\frac{3 u}{3} = \frac{- 3}{3}

Simplify

u = - \frac{3}{3}

u = - \frac{3}{3}

3 u + 3 < 0 : u < - \frac{3}{3}

3 u + 3 < 0

Move

3

to the right side

3 u + 3 < 0

Subtract

3

from both sides

3 u + 3 - 3 < 0 - 3

Simplify

3 u < - 3

3 u < - 3

Divide both sides by

3

3 u < - 3

Divide both sides by

3

\frac{3 u}{3} < \frac{- 3}{3}

Simplify

u < - \frac{3}{3}

u < - \frac{3}{3}

3 u + 3 > 0 : u > - \frac{3}{3}

3 u + 3 > 0

Move

3

to the right side

3 u + 3 > 0

Subtract

3

from both sides

3 u + 3 - 3 > 0 - 3

Simplify

3 u > - 3

3 u > - 3

Divide both sides by

3

3 u > - 3

Divide both sides by

3

\frac{3 u}{3} > \frac{- 3}{3}

Simplify

u > - \frac{3}{3}

u > - \frac{3}{3}

Summarize in a table:

u 3 u + 3 u (3 u + 3) u < - \frac{3}{3} - - + u = - \frac{3}{3} - 00 - \frac{3}{3} < u < 0 - + - u = 0 0 + 0 u > 0 + + +

Identify the intervals that satisfy the required condition:

\leq 0

u = - \frac{3}{3} or - \frac{3}{3} < u < 0 or u = 0

Merge Overlapping Intervals

- \frac{3}{3} \leq u < 0 or u = 0

The union of two intervals is the set of numbers which are in either interval

u = - \frac{3}{3}

- \frac{3}{3} < u < 0

- \frac{3}{3} \leq u < 0

The union of two intervals is the set of numbers which are in either interval

- \frac{3}{3} \leq u < 0

u = 0

- \frac{3}{3} \leq u \leq 0

- \frac{3}{3} \leq u \leq 0

- \frac{3}{3} \leq u \leq 0

- \frac{3}{3} \leq u \leq 0

Substitute back

u = tan (x)

- \frac{3}{3} \leq tan (x) \leq 0

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{3}{3} \leq tan (x) and tan (x) \leq 0

- \frac{3}{3} \leq tan (x) : - \frac{π}{6} + πn \leq x < \frac{π}{2} + πn

- \frac{3}{3} \leq tan (x)

Switch sides

tan (x) \geq - \frac{3}{3}

tan (x) \geq a

then

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

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

tan (x) \leq 0 : - \frac{π}{2} + πn < x \leq πn

tan (x) \leq 0

tan (x) \leq a

then

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

- \frac{π}{2} + πn < x \leq arctan (0) + πn

Simplify

arctan (0) : 0

arctan (0)

Use the following trivial identity:

arctan (0) = 0

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}

= 0

- \frac{π}{2} + πn < x \leq 0 + πn

Simplify

- \frac{π}{2} + πn < x \leq πn

Combine the intervals

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

Merge Overlapping Intervals

\frac{5 π}{6} + πn \leq x < π + πn

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