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Popular Trigonometry >

2sin^2(x)+cos(x)-1>= 0

Solution

2 sin^{2} (x) + cos (x) - 1 \geq 0

Solution

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

Interval Notation

[- \frac{2 π}{3} + 2 πn, \frac{2 π}{3} + 2 πn]

Decimal

- 2.09439 \dots + 2 πn \leq x \leq 2.09439 \dots + 2 πn

Solution steps

2 sin^{2} (x) + cos (x) - 1 \geq 0

Use the following identity:

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

Therefore

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

2 (1 - cos^{2} (x)) + cos (x) - 1 \geq 0

Simplify

2 (1 - cos^{2} (x)) + cos (x) - 1 : cos (x) - 2 cos^{2} (x) + 1

2 (1 - cos^{2} (x)) + cos (x) - 1

Expand

2 (1 - cos^{2} (x)) : 2 - 2 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

= 2 \cdot 1 - 2 cos^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

2 - 2 cos^{2} (x) + cos (x) - 1 : cos (x) - 2 cos^{2} (x) + 1

2 - 2 cos^{2} (x) + cos (x) - 1

Group like terms

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

Add/Subtract the numbers:

2 - 1 = 1

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

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

cos (x) - 2 cos^{2} (x) + 1 \geq 0

Let:

u = cos (x)

u - 2 u^{2} + 1 \geq 0

u - 2 u^{2} + 1 \geq 0 : - \frac{1}{2} \leq u \leq 1

u - 2 u^{2} + 1 \geq 0

Factor

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

u - 2 u^{2} + 1

Factor out common term

- 1

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

Factor

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

2 u^{2} - u - 1

Write in the standard form

a x^{2} + b x + c

= 2 u^{2} - u - 1

Break the expression into groups

2 u^{2} - 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 = - 1

Check

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

TrueCheck

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

False

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)

- (2 u + 1) (u - 1) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (2 u + 1) (u - 1)) (- 1) \leq 0 \cdot (- 1)

Simplify

(2 u + 1) (u - 1) \leq 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

Subtract

1

from 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

Subtract

1

from 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

Subtract

1

from 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:

\leq 0

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

Merge Overlapping Intervals

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

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

u = - \frac{1}{2}

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

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

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

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

u = 1

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

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

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

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

Substitute back

u = cos (x)

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

a \leq u \leq b

then

a \leq u and u \leq b

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

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

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

Switch sides

cos (x) \geq - \frac{1}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

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

Simplify

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

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

Use the following trivial identity:

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

Simplify

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

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

Use the following trivial identity:

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

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

cos (x) \leq 1 :

True for all

x \in R

cos (x) \leq 1

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

Let

y = cos (x)

Combine the intervals

y \leq 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq 1 and - 1 \leq y \leq 1

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

y \leq 1

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{2 π}{3} + 2 πn \leq x \leq \frac{2 π}{3} + 2 πn and True for all x \in R

Merge Overlapping Intervals

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

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