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Popular Trigonometry >

tan^2(x)>= sqrt(3)tan(x)

Solution

tan^{2} (x) \geq 3 tan (x)

Solution

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

Interval Notation

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

Decimal

1.04719 \dots + πn \leq x < 1.57079 \dots + πn or - 1.57079 \dots + πn < x \leq πn

Solution steps

tan^{2} (x) \geq 3 tan (x)

Let:

u = tan (x)

u^{2} \geq 3 u

u^{2} \geq 3 u : u \leq 0 or u \geq 3

u^{2} \geq 3 u

Rewrite in standard form

u^{2} \geq 3 u

Subtract

3 u

from both sides

u^{2} - 3 u \geq 3 u - 3 u

Simplify

u^{2} - 3 u \geq 0

u^{2} - 3 u \geq 0

Factor

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

u^{2} - 3 u

Apply exponent rule:

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

u^{2} = uu

= uu - 3 u

Factor out common term

u

= u (u - 3)

u (u - 3) \geq 0

Identify the intervals

Find the signs of the factors of

u (u - 3)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

u - 3

u - 3 = 0 : u = 3

u - 3 = 0

Move

3

to the right side

u - 3 = 0

Add

3

to both sides

u - 3 + 3 = 0 + 3

Simplify

u = 3

u = 3

u - 3 < 0 : u < 3

u - 3 < 0

Move

3

to the right side

u - 3 < 0

Add

3

to both sides

u - 3 + 3 < 0 + 3

Simplify

u < 3

u < 3

u - 3 > 0 : u > 3

u - 3 > 0

Move

3

to the right side

u - 3 > 0

Add

3

to both sides

u - 3 + 3 > 0 + 3

Simplify

u > 3

u > 3

Summarize in a table:

u u - 3 u (u - 3) u < 0 - - + u = 0 0 - 0 0 < u < 3 + - - u = 3 + 00 u > 3 + + +

Identify the intervals that satisfy the required condition:

\geq 0

u < 0 or u = 0 or u = 3 or u > 3

Merge Overlapping Intervals

u \leq 0 or u = 3 or u > 3

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

u < 0

u = 0

u \leq 0

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

u \leq 0

u = 3

u \leq 0 or u = 3

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

u \leq 0 or u = 3

u > 3

u \leq 0 or u \geq 3

u \leq 0 or u \geq 3

u \leq 0 or u \geq 3

u \leq 0 or u \geq 3

Substitute back

u = tan (x)

tan (x) \leq 0 or tan (x) \geq 3

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

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

tan (x) \geq 3

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 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}{3} + πn \leq x < \frac{π}{2} + πn

Combine the intervals

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

Merge Overlapping Intervals

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

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