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Popular >

tan^2(x)-3tan(x)-4<0

Solution

tan^{2} (x) - 3 tan (x) - 4 < 0

Solution

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

Interval Notation

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

Decimal Notation

πn \leq x < 1.32581 \dots + πn or 2.35619 \dots + πn < x < 3.14159 \dots + πn

Solution steps

tan^{2} (x) - 3 tan (x) - 4 < 0

Let:

u = tan (x)

u^{2} - 3 u - 4 < 0

u^{2} - 3 u - 4 < 0 : - 1 < u < 4

u^{2} - 3 u - 4 < 0

Factor

u^{2} - 3 u - 4 : (u + 1) (u - 4)

u^{2} - 3 u - 4

Break the expression into groups

u^{2} - 3 u - 4

Definition

Factors of

4 : 1, 2, 4

4

Divisors (Factors)

Find the Prime factors of

4 : 2, 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2

Add the prime factors:

2

Add 1 and the number

4

itself

1, 4

The factors of

4

1, 2, 4

Negative factors of

4 : - 1, - 2, - 4

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 4

For every two factors such that

u * v = - 4,

check if

u + v = - 3

Check

u = 1, v = - 4 : u * v = - 4, u + v = - 3 \Rightarrow

TrueCheck

u = 2, v = - 2 : u * v = - 4, u + v = 0 \Rightarrow

False

u = 1, v = - 4

Group into

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

(u^{2} + u) + (- 4 u - 4)

= (u^{2} + u) + (- 4 u - 4)

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

- 4

from

- 4 u - 4 : - 4 (u + 1)

- 4 u - 4

Factor out common term

- 4

= - 4 (u + 1)

= u (u + 1) - 4 (u + 1)

Factor out common term

u + 1

= (u + 1) (u - 4)

(u + 1) (u - 4) < 0

Identify the intervals

Find the signs of the factors of

(u + 1) (u - 4)

Find the signs of

u + 1

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from 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

Subtract

1

from 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

Subtract

1

from both sides

u + 1 - 1 > 0 - 1

Simplify

u > - 1

u > - 1

Find the signs of

u - 4

u - 4 = 0 : u = 4

u - 4 = 0

Move

4

to the right side

u - 4 = 0

Add

4

to both sides

u - 4 + 4 = 0 + 4

Simplify

u = 4

u = 4

u - 4 < 0 : u < 4

u - 4 < 0

Move

4

to the right side

u - 4 < 0

Add

4

to both sides

u - 4 + 4 < 0 + 4

Simplify

u < 4

u < 4

u - 4 > 0 : u > 4

u - 4 > 0

Move

4

to the right side

u - 4 > 0

Add

4

to both sides

u - 4 + 4 > 0 + 4

Simplify

u > 4

u > 4

Summarize in a table:

u + 1 u - 4 (u + 1) (u - 4) u < - 1 - - + u = - 1 0 - 0 - 1 < u < 4 + - - u = 4 + 00 u > 4 + + +

Identify the intervals that satisfy the required condition:

< 0

- 1 < u < 4

- 1 < u < 4

- 1 < u < 4

Substitute back

u = tan (x)

- 1 < tan (x) < 4

a < u < b

then

a < u and u < b

- 1 < tan (x) and tan (x) < 4

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

- 1 < tan (x)

Switch sides

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 property:

arctan (- x) = - arctan (x)

arctan (- 1) = - arctan (1)

= - arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

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}

- \frac{π}{4} + πn < x < \frac{π}{2} + πn

tan (x) < 4 : - \frac{π}{2} + πn < x < arctan (4) + πn

tan (x) < 4

tan (x) < a

then

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

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

Combine the intervals

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

Merge Overlapping Intervals

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

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