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Popular Trigonometry >

cos^2(x)-4cos(x)<0

Solution

cos^{2} (x) - 4 cos (x) < 0

Solution

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

Interval Notation

(- \frac{π}{2} + 2 πn, \frac{π}{2} + 2 πn)

Decimal

- 1.57079 \dots + 2 πn < x < 1.57079 \dots + 2 πn

Solution steps

cos^{2} (x) - 4 cos (x) < 0

Let:

u = cos (x)

u^{2} - 4 u < 0

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

u^{2} - 4 u < 0

Factor

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

u^{2} - 4 u

Apply exponent rule:

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

u^{2} = uu

= uu - 4 u

Factor out common term

u

= u (u - 4)

u (u - 4) < 0

Identify the intervals

Find the signs of the factors of

u (u - 4)

Find the signs of

u

u = 0

u < 0

u > 0

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

Identify the intervals that satisfy the required condition:

< 0

0 < u < 4

0 < u < 4

0 < u < 4

Substitute back

u = cos (x)

0 < cos (x) < 4

a < u < b

then

a < u and u < b

0 < cos (x) and cos (x) < 4

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

0 < cos (x)

Switch sides

cos (x) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < x < arccos (0) + 2 πn

Simplify

- arccos (0) : - \frac{π}{2}

- arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2}

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

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

cos (x) < 4 :

True for all

x \in R

cos (x) < 4

Range of

cos (x) : - 1 \leq cos (x) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (x) \leq 1

- 1 \leq cos (x) \leq 1

cos (x) < 4 and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

Let

y = cos (x)

Combine the intervals

y < 4 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < 4 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y < 4

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

Combine the intervals

- \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn and True for all x \in R

Merge Overlapping Intervals

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

Popular Examples

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