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Popular Trigonometry >

0<sin(x+pi/3)<2pi

Solution

0 < sin (x + \frac{π}{3}) < 2 π

Solution

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

Interval Notation

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

Decimal

- 1.04719 \dots + 2 πn < x < 2.09439 \dots + 2 πn

Solution steps

0 < sin (x + \frac{π}{3}) < 2 π

a < u < b

then

a < u and u < b

0 < sin (x + \frac{π}{3}) and sin (x + \frac{π}{3}) < 2 π

0 < sin (x + \frac{π}{3}) : - \frac{π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn

0 < sin (x + \frac{π}{3})

Switch sides

sin (x + \frac{π}{3}) > 0

For

sin (x) > a

, if

- 1 \leq a < 1

then

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

arcsin (0) + 2 πn < (x + \frac{π}{3}) < π - arcsin (0) + 2 πn

a < u < b

then

a < u and u < b

arcsin (0) + 2 πn < x + \frac{π}{3} and x + \frac{π}{3} < π - arcsin (0) + 2 πn

arcsin (0) + 2 πn < x + \frac{π}{3} : x > 2 πn - \frac{π}{3}

arcsin (0) + 2 πn < x + \frac{π}{3}

Switch sides

x + \frac{π}{3} > arcsin (0) + 2 πn

Simplify

arcsin (0) + 2 πn : 2 πn

arcsin (0) + 2 πn

Use the following trivial identity:

arcsin (0) = 0

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}

= 0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

x + \frac{π}{3} > 2 πn

Move

\frac{π}{3}

to the right side

x + \frac{π}{3} > 2 πn

Subtract

\frac{π}{3}

from both sides

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

Simplify

x > 2 πn - \frac{π}{3}

x > 2 πn - \frac{π}{3}

x + \frac{π}{3} < π - arcsin (0) + 2 πn : x < \frac{2 π}{3} + 2 πn

x + \frac{π}{3} < π - arcsin (0) + 2 πn

Simplify

π - arcsin (0) + 2 πn : π + 2 πn

π - arcsin (0) + 2 πn

Use the following trivial identity:

arcsin (0) = 0

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 - 0 + 2 πn

π - 0 + 2 πn = π + 2 πn

= π + 2 πn

x + \frac{π}{3} < π + 2 πn

Move

\frac{π}{3}

to the right side

x + \frac{π}{3} < π + 2 πn

Subtract

\frac{π}{3}

from both sides

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

Simplify

x < π + 2 πn - \frac{π}{3}

x < π + 2 πn - \frac{π}{3}

Simplify

π - \frac{π}{3} : \frac{2 π}{3}

π - \frac{π}{3}

Convert element to fraction:

π = \frac{π 3}{3}

= \frac{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{π 3 - π}{3}

Add similar elements:

3 π - π = 2 π

= \frac{2 π}{3}

x < \frac{2 π}{3} + 2 πn

Combine the intervals

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

Merge Overlapping Intervals

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

sin (x + \frac{π}{3}) < 2 π :

True for all

x \in R

sin (x + \frac{π}{3}) < 2 π

Range of

sin (x + \frac{π}{3}) : - 1 \leq sin (x + \frac{π}{3}) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x + \frac{π}{3}) \leq 1

- 1 \leq sin (x + \frac{π}{3}) \leq 1

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

Let

y = sin (x + \frac{π}{3})

Combine the intervals

y < 2 π and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < 2 π and - 1 \leq y \leq 1

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

y < 2 π

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

Merge Overlapping Intervals

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

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