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Popular >

tan^3(x)<=-sqrt(3)tan(x)

Solution

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

Solution

- \frac{π}{2} + πn < x \leq πn

Interval Notation

(- \frac{π}{2} + πn, πn]

Decimal Notation

- 1.57079 \dots + πn < x \leq πn

Solution steps

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

Let:

u = tan (x)

u^{3} \leq - 3 u

u^{3} \leq - 3 u : u \leq 0

u^{3} \leq - 3 u

Rewrite in standard form

u^{3} \leq - 3 u

Add

3 u

to both sides

u^{3} + 3 u \leq - 3 u + 3 u

Simplify

u^{3} + 3 u \leq 0

u^{3} + 3 u \leq 0

Factor

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

u^{3} + 3 u

Apply exponent rule:

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

u^{3} = u^{2} u

= u^{2} u + 3 u

Factor out common term

u

= u (u^{2} + 3)

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

Identify the intervals

Find the signs of the factors of

u (u^{2} + 3)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

u^{2} + 3 :

Positive for all real values

u^{2} + 3 > 0

Identify the intervals

Find the signs of the factors of

u^{2} + 3

Find the signs of

u^{2} + 3

Summarize in a table:

u^{2} + 3 u^{2} + 3 - \infty < u < \infty + +

Identify the intervals that satisfy the required condition:

> 0

True for all u \in R

True for all u

True for all u \in R

Summarize in a table:

u u^{2} + 3 u (u^{2} + 3) u < 0 - + - u = 0 0 + 0 u > 0 + + +

Identify the intervals that satisfy the required condition:

\leq 0

u < 0 or u = 0

Merge Overlapping Intervals

u < 0 or u = 0

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

u < 0

u = 0

u \leq 0

u \leq 0

u \leq 0

u \leq 0

Substitute back

u = tan (x)

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

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