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Popular >

2cos^2(x)-3cos(x)+1<0

Solution

2 cos^{2} (x) - 3 cos (x) + 1 < 0

Solution

2 πn < x < \frac{π}{3} + 2 πn or \frac{5 π}{3} + 2 πn < x < 2 π + 2 πn

Interval Notation

(2 πn, \frac{π}{3} + 2 πn) \cup (\frac{5 π}{3} + 2 πn, 2 π + 2 πn)

Decimal Notation

2 πn < x < 1.04719 \dots + 2 πn or 5.23598 \dots + 2 πn < x < 6.28318 \dots + 2 πn

Solution steps

2 cos^{2} (x) - 3 cos (x) + 1 < 0

Let:

u = cos (x)

2 u^{2} - 3 u + 1 < 0

2 u^{2} - 3 u + 1 < 0 : \frac{1}{2} < u < 1

2 u^{2} - 3 u + 1 < 0

Factor

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

2 u^{2} - 3 u + 1

Break the expression into groups

2 u^{2} - 3 u + 1

Definition

Factors of

2 : 1, 2

2

Divisors (Factors)

Find the Prime factors of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Add 1

1

The factors of

2

1, 2

Negative factors of

2 : - 1, - 2

Multiply the factors by

- 1

to get the negative factors

- 1, - 2

For every two factors such that

u * v = 2,

check if

u + v = - 3

Check

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

FalseCheck

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

True

u = - 1, v = - 2

Group into

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

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

= (2 u^{2} - u) + (- 2 u + 1)

Factor out

u

from

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

2 u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= 2 uu - u

Factor out common term

u

= u (2 u - 1)

Factor out

- 1

from

- 2 u + 1 : - (2 u - 1)

- 2 u + 1

Factor out common term

- 1

= - (2 u - 1)

= u (2 u - 1) - (2 u - 1)

Factor out common term

2 u - 1

= (2 u - 1) (u - 1)

(2 u - 1) (u - 1) < 0

Identify the intervals

Find the signs of the factors of

(2 u - 1) (u - 1)

Find the signs of

2 u - 1

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

2 u - 1 < 0 : u < \frac{1}{2}

2 u - 1 < 0

Move

1

to the right side

2 u - 1 < 0

Add

1

to both sides

2 u - 1 + 1 < 0 + 1

Simplify

2 u < 1

2 u < 1

Divide both sides by

2

2 u < 1

Divide both sides by

2

\frac{2 u}{2} < \frac{1}{2}

Simplify

u < \frac{1}{2}

u < \frac{1}{2}

2 u - 1 > 0 : u > \frac{1}{2}

2 u - 1 > 0

Move

1

to the right side

2 u - 1 > 0

Add

1

to both sides

2 u - 1 + 1 > 0 + 1

Simplify

2 u > 1

2 u > 1

Divide both sides by

2

2 u > 1

Divide both sides by

2

\frac{2 u}{2} > \frac{1}{2}

Simplify

u > \frac{1}{2}

u > \frac{1}{2}

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

Summarize in a table:

2 u - 1 u - 1 (2 u - 1) (u - 1) u < \frac{1}{2} - - + u = \frac{1}{2} 0 - 0 \frac{1}{2} < u < 1 + - - u = 1 + 00 u > 1 + + +

Identify the intervals that satisfy the required condition:

< 0

\frac{1}{2} < u < 1

\frac{1}{2} < u < 1

\frac{1}{2} < u < 1

Substitute back

u = cos (x)

\frac{1}{2} < cos (x) < 1

a < u < b

then

a < u and u < b

\frac{1}{2} < cos (x) and cos (x) < 1

\frac{1}{2} < cos (x) : - \frac{π}{3} + 2 πn < x < \frac{π}{3} + 2 πn

\frac{1}{2} < cos (x)

Switch sides

cos (x) > \frac{1}{2}

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

- arccos (\frac{1}{2}) + 2 πn < x < arccos (\frac{1}{2}) + 2 πn

Simplify

- arccos (\frac{1}{2}) : - \frac{π}{3}

- arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3}

Simplify

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3}

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

cos (x) < 1 : 2 πn < x < 2 π + 2 πn

cos (x) < 1

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

arccos (1) + 2 πn < x < 2 π - arccos (1) + 2 πn

Simplify

arccos (1) : 0

arccos (1)

Use the following trivial identity:

arccos (1) = 0

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}

= 0

Simplify

2 π - arccos (1) : 2 π

2 π - arccos (1)

Use the following trivial identity:

arccos (1) = 0

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}

= 2 π - 0

2 π - 0 = 2 π

= 2 π

0 + 2 πn < x < 2 π + 2 πn

Simplify

2 πn < x < 2 π + 2 πn

Combine the intervals

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

Merge Overlapping Intervals

2 πn < x < \frac{π}{3} + 2 πn or \frac{5 π}{3} + 2 πn < x < 2 π + 2 πn

Popular Examples

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sqrt(3)cot(x/2)>-3

3 cot (\frac{x}{2}) > - 3

tan(θ)>0,cos(θ)<0sin(θ)=-2/5 ,tan(θ)

tan (θ) > 0, cos (θ) < 0 sin (θ) = - \frac{2}{5}, tan (θ)

tan^2(x)>1

tan^{2} (x) > 1