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Popular Trigonometry >

7cos^2(x)-5cos(x)+sin^2(x)<= 0

Solution

7 cos^{2} (x) - 5 cos (x) + sin^{2} (x) \leq 0

Solution

\frac{π}{3} + 2 πn \leq x \leq arccos (\frac{1}{3}) + 2 πn or - arccos (\frac{1}{3}) + 2 π + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

Interval Notation

[\frac{π}{3} + 2 πn, arccos (\frac{1}{3}) + 2 πn] \cup [- arccos (\frac{1}{3}) + 2 π + 2 πn, \frac{5 π}{3} + 2 πn]

Decimal

1.04719 \dots + 2 πn \leq x \leq 1.23095 \dots + 2 πn or 5.05222 \dots + 2 πn \leq x \leq 5.23598 \dots + 2 πn

Solution steps

7 cos^{2} (x) - 5 cos (x) + sin^{2} (x) \leq 0

Use the following identity:

cos^{2} (x) + sin^{2} (x) = 1

Therefore

sin^{2} (x) = 1 - cos^{2} (x)

7 cos^{2} (x) - 5 cos (x) + 1 - cos^{2} (x) \leq 0

Simplify

6 cos^{2} (x) - 5 cos (x) + 1 \leq 0

Let:

u = cos (x)

6 u^{2} - 5 u + 1 \leq 0

6 u^{2} - 5 u + 1 \leq 0 : \frac{1}{3} \leq u \leq \frac{1}{2}

6 u^{2} - 5 u + 1 \leq 0

Factor

6 u^{2} - 5 u + 1 : (3 u - 1) (2 u - 1)

6 u^{2} - 5 u + 1

Break the expression into groups

6 u^{2} - 5 u + 1

Definition

Factors of

6 : 1, 2, 3, 6

6

Divisors (Factors)

Find the Prime factors of

6 : 2, 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Add the prime factors:

2, 3

Add 1 and the number

6

itself

1, 6

The factors of

6

1, 2, 3, 6

Negative factors of

6 : - 1, - 2, - 3, - 6

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 6

For every two factors such that

u * v = 6,

check if

u + v = - 5

Check

u = 1, v = 6 : u * v = 6, u + v = 7 \Rightarrow

FalseCheck

u = 2, v = 3 : u * v = 6, u + v = 5 \Rightarrow

False

u = - 2, v = - 3

Group into

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

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

= (6 u^{2} - 2 u) + (- 3 u + 1)

Factor out

2 u

from

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

6 u^{2} - 2 u

Apply exponent rule:

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

u^{2} = uu

= 6 uu - 2 u

Rewrite

6

2 \cdot 3

= 2 \cdot 3 uu - 2 u

Factor out common term

2 u

= 2 u (3 u - 1)

Factor out

- 1

from

- 3 u + 1 : - (3 u - 1)

- 3 u + 1

Factor out common term

- 1

= - (3 u - 1)

= 2 u (3 u - 1) - (3 u - 1)

Factor out common term

3 u - 1

= (3 u - 1) (2 u - 1)

(3 u - 1) (2 u - 1) \leq 0

Identify the intervals

Find the signs of the factors of

(3 u - 1) (2 u - 1)

Find the signs of

3 u - 1

3 u - 1 = 0 : u = \frac{1}{3}

3 u - 1 = 0

Move

1

to the right side

3 u - 1 = 0

Add

1

to both sides

3 u - 1 + 1 = 0 + 1

Simplify

3 u = 1

3 u = 1

Divide both sides by

3

3 u = 1

Divide both sides by

3

\frac{3 u}{3} = \frac{1}{3}

Simplify

u = \frac{1}{3}

u = \frac{1}{3}

3 u - 1 < 0 : u < \frac{1}{3}

3 u - 1 < 0

Move

1

to the right side

3 u - 1 < 0

Add

1

to both sides

3 u - 1 + 1 < 0 + 1

Simplify

3 u < 1

3 u < 1

Divide both sides by

3

3 u < 1

Divide both sides by

3

\frac{3 u}{3} < \frac{1}{3}

Simplify

u < \frac{1}{3}

u < \frac{1}{3}

3 u - 1 > 0 : u > \frac{1}{3}

3 u - 1 > 0

Move

1

to the right side

3 u - 1 > 0

Add

1

to both sides

3 u - 1 + 1 > 0 + 1

Simplify

3 u > 1

3 u > 1

Divide both sides by

3

3 u > 1

Divide both sides by

3

\frac{3 u}{3} > \frac{1}{3}

Simplify

u > \frac{1}{3}

u > \frac{1}{3}

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}

Summarize in a table:

3 u - 1 2 u - 1 (3 u - 1) (2 u - 1) u < \frac{1}{3} - - + u = \frac{1}{3} 0 - 0 \frac{1}{3} < u < \frac{1}{2} + - - u = \frac{1}{2} + 00 u > \frac{1}{2} + + +

Identify the intervals that satisfy the required condition:

\leq 0

u = \frac{1}{3} or \frac{1}{3} < u < \frac{1}{2} or u = \frac{1}{2}

Merge Overlapping Intervals

\frac{1}{3} \leq u < \frac{1}{2} or u = \frac{1}{2}

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

u = \frac{1}{3}

\frac{1}{3} < u < \frac{1}{2}

\frac{1}{3} \leq u < \frac{1}{2}

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

\frac{1}{3} \leq u < \frac{1}{2}

u = \frac{1}{2}

\frac{1}{3} \leq u \leq \frac{1}{2}

\frac{1}{3} \leq u \leq \frac{1}{2}

\frac{1}{3} \leq u \leq \frac{1}{2}

\frac{1}{3} \leq u \leq \frac{1}{2}

Substitute back

u = cos (x)

\frac{1}{3} \leq cos (x) \leq \frac{1}{2}

a \leq u \leq b

then

a \leq u and u \leq b

\frac{1}{3} \leq cos (x) and cos (x) \leq \frac{1}{2}

\frac{1}{3} \leq cos (x) : - arccos (\frac{1}{3}) + 2 πn \leq x \leq arccos (\frac{1}{3}) + 2 πn

\frac{1}{3} \leq cos (x)

Switch sides

cos (x) \geq \frac{1}{3}

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (\frac{1}{3}) + 2 πn \leq x \leq arccos (\frac{1}{3}) + 2 πn

cos (x) \leq \frac{1}{2} : \frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

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

For

cos (x) \leq a

, if

- 1 < a < 1

then

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

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

2 π - arccos (\frac{1}{2}) : \frac{5 π}{3}

2 π - 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}

= 2 π - \frac{π}{3}

Simplify

2 π - \frac{π}{3}

Convert element to fraction:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{π}{3}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 π 3 - π}{3}

2 π 3 - π = 5 π

2 π 3 - π

Multiply the numbers:

2 \cdot 3 = 6

= 6 π - π

Add similar elements:

6 π - π = 5 π

= 5 π

= \frac{5 π}{3}

= \frac{5 π}{3}

\frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

Combine the intervals

- arccos (\frac{1}{3}) + 2 πn \leq x \leq arccos (\frac{1}{3}) + 2 πn and \frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

Merge Overlapping Intervals

\frac{π}{3} + 2 πn \leq x \leq arccos (\frac{1}{3}) + 2 πn or - arccos (\frac{1}{3}) + 2 π + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

Popular Examples

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cos (- x) < 0

sin(x^2)>= 0

sin (x^{2}) \geq 0

sin(x)+cos(x)<= 2

sin (x) + cos (x) \leq 2

2cos^2(x)+sin(2x)<= 0

2 cos^{2} (x) + sin (2 x) \leq 0

cos(2x)> 1/(sqrt(2))

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