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Popular Trigonometry >

0<= 2sin(3x)+1<2pi

Solution

0 \leq 2 sin (3 x) + 1 < 2 π

Solution

- \frac{π}{18} + \frac{2 π}{3} n \leq x \leq \frac{7 π}{18} + \frac{2 π}{3} n

Interval Notation

[- \frac{π}{18} + \frac{2 π}{3} n, \frac{7 π}{18} + \frac{2 π}{3} n]

Decimal

- 0.17453 \dots + \frac{2 π}{3} n \leq x \leq 1.22173 \dots + \frac{2 π}{3} n

Solution steps

0 \leq 2 sin (3 x) + 1 < 2 π

a \leq u < b

then

a \leq u and u < b

0 \leq 2 sin (3 x) + 1 and 2 sin (3 x) + 1 < 2 π

0 \leq 2 sin (3 x) + 1 : - \frac{π}{18} + \frac{2 π}{3} n \leq x \leq \frac{7 π}{18} + \frac{2 π}{3} n

0 \leq 2 sin (3 x) + 1

Switch sides

2 sin (3 x) + 1 \geq 0

Move

1

to the right side

2 sin (3 x) + 1 \geq 0

Subtract

1

from both sides

2 sin (3 x) + 1 - 1 \geq 0 - 1

Simplify

2 sin (3 x) \geq - 1

2 sin (3 x) \geq - 1

Divide both sides by

2

2 sin (3 x) \geq - 1

Divide both sides by

2

\frac{2 sin ( 3 x )}{2} \geq \frac{- 1}{2}

Simplify

sin (3 x) \geq - \frac{1}{2}

sin (3 x) \geq - \frac{1}{2}

For

sin (x) \geq a

, if

- 1 < a < 1

then

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

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

a \leq u \leq b

then

a \leq u and u \leq b

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

arcsin (- \frac{1}{2}) + 2 πn \leq 3 x : x \geq - \frac{π}{18} + \frac{2 πn}{3}

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

Switch sides

3 x \geq arcsin (- \frac{1}{2}) + 2 πn

Simplify

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

arcsin (- \frac{1}{2}) + 2 πn

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

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

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

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}

= - \frac{π}{6}

= - \frac{π}{6} + 2 πn

3 x \geq - \frac{π}{6} + 2 πn

Divide both sides by

3

3 x \geq - \frac{π}{6} + 2 πn

Divide both sides by

3

\frac{3 x}{3} \geq - \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} \geq - \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

- \frac{\frac{π}{6}}{3} + \frac{2 πn}{3} : - \frac{π}{18} + \frac{2 πn}{3}

- \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= - \frac{π}{18} + \frac{2 πn}{3}

x \geq - \frac{π}{18} + \frac{2 πn}{3}

x \geq - \frac{π}{18} + \frac{2 πn}{3}

x \geq - \frac{π}{18} + \frac{2 πn}{3}

3 x \leq π - arcsin (- \frac{1}{2}) + 2 πn : x \leq \frac{7 π}{18} + \frac{2 π}{3} n

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

Simplify

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

π - arcsin (- \frac{1}{2}) + 2 πn

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

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

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

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}

= - \frac{π}{6}

= π - (- \frac{π}{6}) + 2 πn

Apply rule

- (- a) = a

= π + \frac{π}{6} + 2 πn

3 x \leq π + \frac{π}{6} + 2 πn

Divide both sides by

3

3 x \leq π + \frac{π}{6} + 2 πn

Divide both sides by

3

\frac{3 x}{3} \leq \frac{π}{3} + \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} \leq \frac{π}{3} + \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{π}{3} + \frac{\frac{π}{6}}{3} + \frac{2 πn}{3} : \frac{π}{3} + \frac{π}{18} + \frac{2 πn}{3}

\frac{π}{3} + \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= \frac{π}{3} + \frac{π}{18} + \frac{2 πn}{3}

x \leq \frac{π}{3} + \frac{π}{18} + \frac{2 πn}{3}

x \leq \frac{π}{3} + \frac{π}{18} + \frac{2 πn}{3}

Simplify

\frac{π}{3} + \frac{π}{18} : \frac{7 π}{18}

\frac{π}{3} + \frac{π}{18}

Least Common Multiplier of

3, 18 : 18

3, 18

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

18 : 2 \cdot 3 \cdot 3

18

18

divides by

218 = 9 \cdot 2

= 2 \cdot 9

9

divides by

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 3

Multiply each factor the greatest number of times it occurs in either

3

18

= 3 \cdot 3 \cdot 2

Multiply the numbers:

3 \cdot 3 \cdot 2 = 18

= 18

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

18

For

\frac{π}{3} :

multiply the denominator and numerator by

6

\frac{π}{3} = \frac{π 6}{3 \cdot 6} = \frac{π 6}{18}

= \frac{π 6}{18} + \frac{π}{18}

Since the denominators are equal, combine the fractions:

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

= \frac{π 6 + π}{18}

Add similar elements:

6 π + π = 7 π

= \frac{7 π}{18}

x \leq \frac{7 π}{18} + \frac{2 π}{3} n

x \leq \frac{7 π}{18} + \frac{2 π}{3} n

Combine the intervals

x \geq - \frac{π}{18} + \frac{2 πn}{3} and x \leq \frac{7 π}{18} + \frac{2 π}{3} n

Merge Overlapping Intervals

- \frac{π}{18} + \frac{2 π}{3} n \leq x \leq \frac{7 π}{18} + \frac{2 π}{3} n

2 sin (3 x) + 1 < 2 π :

True for all

x \in R

2 sin (3 x) + 1 < 2 π

Move

1

to the right side

2 sin (3 x) + 1 < 2 π

Subtract

1

from both sides

2 sin (3 x) + 1 - 1 < 2 π - 1

Simplify

2 sin (3 x) < 2 π - 1

2 sin (3 x) < 2 π - 1

Divide both sides by

2

2 sin (3 x) < 2 π - 1

Divide both sides by

2

\frac{2 sin ( 3 x )}{2} < \frac{2 π}{2} - \frac{1}{2}

Simplify

\frac{2 sin ( 3 x )}{2} < \frac{2 π}{2} - \frac{1}{2}

Simplify

\frac{2 sin ( 3 x )}{2} : sin (3 x)

\frac{2 sin ( 3 x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin (3 x)

Simplify

\frac{2 π}{2} - \frac{1}{2} : \frac{2 π - 1}{2}

\frac{2 π}{2} - \frac{1}{2}

Apply rule

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

= \frac{2 π - 1}{2}

sin (3 x) < \frac{2 π - 1}{2}

sin (3 x) < \frac{2 π - 1}{2}

sin (3 x) < \frac{2 π - 1}{2}

Range of

sin (3 x) : - 1 \leq sin (3 x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (3 x) \leq 1

- 1 \leq sin (3 x) \leq 1

sin (3 x) < \frac{2 π - 1}{2} and - 1 \leq sin (3 x) \leq 1 : - 1 \leq sin (3 x) \leq 1

Let

y = sin (3 x)

Combine the intervals

y < \frac{2 π - 1}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < \frac{2 π - 1}{2} and - 1 \leq y \leq 1

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

y < \frac{2 π - 1}{2}

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}{18} + \frac{2 π}{3} n \leq x \leq \frac{7 π}{18} + \frac{2 π}{3} n and True for all x \in R

Merge Overlapping Intervals

- \frac{π}{18} + \frac{2 π}{3} n \leq x \leq \frac{7 π}{18} + \frac{2 π}{3} n

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