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Popular Trigonometry >

tan^2(x)+2tan(x)>3

Solution

tan^{2} (x) + 2 tan (x) > 3

Solution

\frac{π}{4} + πn < x < \frac{π}{2} + πn or - \frac{π}{2} + πn < x < - arctan (3) + πn

Interval Notation

(\frac{π}{4} + πn, \frac{π}{2} + πn) \cup (- \frac{π}{2} + πn, - arctan (3) + πn)

Decimal

0.78539 \dots + πn < x < 1.57079 \dots + πn or - 1.57079 \dots + πn < x < - 1.24904 \dots + πn

Solution steps

tan^{2} (x) + 2 tan (x) > 3

Let:

u = tan (x)

u^{2} + 2 u > 3

u^{2} + 2 u > 3 : u < - 3 or u > 1

u^{2} + 2 u > 3

Rewrite in standard form

u^{2} + 2 u > 3

Subtract

3

from both sides

u^{2} + 2 u - 3 > 3 - 3

Simplify

u^{2} + 2 u - 3 > 0

u^{2} + 2 u - 3 > 0

Factor

u^{2} + 2 u - 3 : (u - 1) (u + 3)

u^{2} + 2 u - 3

Break the expression into groups

u^{2} + 2 u - 3

Definition

Factors of

3 : 1, 3

3

Divisors (Factors)

Find the Prime factors of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Add 1

1

The factors of

3

1, 3

Negative factors of

3 : - 1, - 3

Multiply the factors by

- 1

to get the negative factors

- 1, - 3

For every two factors such that

u * v = - 3,

check if

u + v = 2

Check

u = 1, v = - 3 : u * v = - 3, u + v = - 2 \Rightarrow

FalseCheck

u = 3, v = - 1 : u * v = - 3, u + v = 2 \Rightarrow

True

u = 3, v = - 1

Group into

(a x^{2} + ux) + (vx + c)

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

= (u^{2} - u) + (3 u - 3)

Factor out

u

from

u^{2} - u : u (u - 1)

u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= uu - u

Factor out common term

u

= u (u - 1)

Factor out

3

from

3 u - 3 : 3 (u - 1)

3 u - 3

Factor out common term

3

= 3 (u - 1)

= u (u - 1) + 3 (u - 1)

Factor out common term

u - 1

= (u - 1) (u + 3)

(u - 1) (u + 3) > 0

Identify the intervals

Find the signs of the factors of

(u - 1) (u + 3)

Find the signs of

u - 1

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

u - 1 < 0 : u < 1

u - 1 < 0

Move

1

to the right side

u - 1 < 0

Add

1

to both sides

u - 1 + 1 < 0 + 1

Simplify

u < 1

u < 1

u - 1 > 0 : u > 1

u - 1 > 0

Move

1

to the right side

u - 1 > 0

Add

1

to both sides

u - 1 + 1 > 0 + 1

Simplify

u > 1

u > 1

Find the signs of

u + 3

u + 3 = 0 : u = - 3

u + 3 = 0

Move

3

to the right side

u + 3 = 0

Subtract

3

from 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

Subtract

3

from 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

Subtract

3

from both sides

u + 3 - 3 > 0 - 3

Simplify

u > - 3

u > - 3

Summarize in a table:

u - 1 u + 3 (u - 1) (u + 3) u < - 3 - - + u = - 3 - 00 - 3 < u < 1 - + - u = 1 0 + 0 u > 1 + + +

Identify the intervals that satisfy the required condition:

> 0

u < - 3 or u > 1

u < - 3 or u > 1

u < - 3 or u > 1

Substitute back

u = tan (x)

tan (x) < - 3 or tan (x) > 1

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

tan (x) < - 3

tan (x) < a

then

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

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

Simplify

arctan (- 3) : - arctan (3)

arctan (- 3)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 3) = - arctan (3)

= - arctan (3)

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

tan (x) > 1 : \frac{π}{4} + πn < x < \frac{π}{2} + πn

tan (x) > 1

tan (x) > a

then

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

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

Simplify

arctan (1) : \frac{π}{4}

arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

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} + πn < x < \frac{π}{2} + πn

Combine the intervals

- \frac{π}{2} + πn < x < - arctan (3) + πn or \frac{π}{4} + πn < x < \frac{π}{2} + πn

Merge Overlapping Intervals

\frac{π}{4} + πn < x < \frac{π}{2} + πn or - \frac{π}{2} + πn < x < - arctan (3) + πn

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