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Popular >

-12cos(2x)+12sin(x)>0

Solution

- 12 cos (2 x) + 12 sin (x) > 0

Solution

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn

Interval Notation

(\frac{π}{6} + 2 πn, \frac{5 π}{6} + 2 πn)

Decimal Notation

0.52359 \dots + 2 πn < x < 2.61799 \dots + 2 πn

Solution steps

- 12 cos (2 x) + 12 sin (x) > 0

Use the following identity:

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

- 12 (1 - 2 sin^{2} (x)) + 12 sin (x) > 0

Let:

u = sin (x)

- 12 (1 - 2 u^{2}) + 12 u > 0

- 12 (1 - 2 u^{2}) + 12 u > 0 : u < - 1 or u > \frac{1}{2}

- 12 (1 - 2 u^{2}) + 12 u > 0

Rewrite in standard form

- 12 (1 - 2 u^{2}) + 12 u > 0

Expand

- 12 (1 - 2 u^{2}) + 12 u : - 12 + 24 u^{2} + 12 u

- 12 (1 - 2 u^{2}) + 12 u

Expand

- 12 (1 - 2 u^{2}) : - 12 + 24 u^{2}

- 12 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 12, b = 1, c = 2 u^{2}

= - 12 \cdot 1 - (- 12) \cdot 2 u^{2}

Apply minus-plus rules

- (- a) = a

= - 12 \cdot 1 + 12 \cdot 2 u^{2}

Simplify

- 12 \cdot 1 + 12 \cdot 2 u^{2} : - 12 + 24 u^{2}

- 12 \cdot 1 + 12 \cdot 2 u^{2}

Multiply the numbers:

12 \cdot 1 = 12

= - 12 + 12 \cdot 2 u^{2}

Multiply the numbers:

12 \cdot 2 = 24

= - 12 + 24 u^{2}

= - 12 + 24 u^{2}

= - 12 + 24 u^{2} + 12 u

- 12 + 24 u^{2} + 12 u > 0

Divide both sides by

12

- \frac{12}{12} + \frac{24 u ^{2}}{12} + \frac{12 u}{12} > \frac{0}{12}

Refine

- \frac{12}{12} + \frac{24 u ^{2}}{12} + \frac{12 u}{12} > \frac{0}{12} : 2 u^{2} + u - 1 > 0

- \frac{12}{12} + \frac{24 u ^{2}}{12} + \frac{12 u}{12} > \frac{0}{12}

Simplify

- \frac{12}{12} + \frac{24 u ^{2}}{12} + \frac{12 u}{12} : 2 u^{2} + u - 1

- \frac{12}{12} + \frac{24 u ^{2}}{12} + \frac{12 u}{12}

Apply rule

\frac{a}{a} = 1

\frac{12}{12} = 1

= - 1 + \frac{24 u ^{2}}{12} + \frac{12 u}{12}

Divide the numbers:

\frac{24}{12} = 2

= - 1 + 2 u^{2} + \frac{12 u}{12}

Divide the numbers:

\frac{12}{12} = 1

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

Rewrite in standard form

= 2 u^{2} + u - 1

\frac{0}{12} = 0

\frac{0}{12}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

2 u^{2} + u - 1 > 0

2 u^{2} + u - 1 > 0

2 u^{2} + u - 1 > 0

Factor

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

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

FalseCheck

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

True

u = 2, v = - 1

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)

= 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

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

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

> 0

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

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

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

Substitute back

u = sin (x)

sin (x) < - 1 or sin (x) > \frac{1}{2}

sin (x) < - 1 :

False for all

x \in R

sin (x) < - 1

Range of

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

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) < - 1 and - 1 \leq sin (x) \leq 1 :

False

Let

y = sin (x)

Combine the intervals

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

Merge Overlapping Intervals

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

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

y < - 1

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

sin (x) > \frac{1}{2} : \frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn

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

For

sin (x) > a

, if

- 1 \leq a < 1

then

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

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

Simplify

arcsin (\frac{1}{2}) : \frac{π}{6}

arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

Simplify

π - arcsin (\frac{1}{2}) : \frac{5 π}{6}

π - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= π - \frac{π}{6}

Simplify

π - \frac{π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= \frac{π 6}{6} - \frac{π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π 6 - π}{6}

Add similar elements:

6 π - π = 5 π

= \frac{5 π}{6}

= \frac{5 π}{6}

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn

Combine the intervals

False for all x \in R or \frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn

Merge Overlapping Intervals

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn

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