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Popular Trigonometry >

sin(x+pi/4)<= 1/2

Solution

sin (x + \frac{π}{4}) \leq \frac{1}{2}

Solution

- \frac{17 π}{12} + 2 πn \leq x \leq - \frac{π}{12} + 2 πn

Interval Notation

[- \frac{17 π}{12} + 2 πn, - \frac{π}{12} + 2 πn]

Decimal

- 4.45058 \dots + 2 πn \leq x \leq - 0.26179 \dots + 2 πn

Solution steps

sin (x + \frac{π}{4}) \leq \frac{1}{2}

For

sin (x) \leq a

, if

- 1 < a < 1

then

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

- π - arcsin (\frac{1}{2}) + 2 πn \leq (x + \frac{π}{4}) \leq arcsin (\frac{1}{2}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- π - arcsin (\frac{1}{2}) + 2 πn \leq x + \frac{π}{4} and x + \frac{π}{4} \leq arcsin (\frac{1}{2}) + 2 πn

- π - arcsin (\frac{1}{2}) + 2 πn \leq x + \frac{π}{4} : x \geq - \frac{17 π}{12} + 2 πn

- π - arcsin (\frac{1}{2}) + 2 πn \leq x + \frac{π}{4}

Switch sides

x + \frac{π}{4} \geq - 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

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} + 2 πn

x + \frac{π}{4} \geq - π - \frac{π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

x + \frac{π}{4} \geq - π - \frac{π}{6} + 2 πn

Subtract

\frac{π}{4}

from both sides

x + \frac{π}{4} - \frac{π}{4} \geq - π - \frac{π}{6} + 2 πn - \frac{π}{4}

Simplify

x + \frac{π}{4} - \frac{π}{4} \geq - π - \frac{π}{6} + 2 πn - \frac{π}{4}

Simplify

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} \geq 0

= x

Simplify

- π - \frac{π}{6} + 2 πn - \frac{π}{4} : - π + 2 πn - \frac{5 π}{12}

- π - \frac{π}{6} + 2 πn - \frac{π}{4}

Group like terms

= - π + 2 πn - \frac{π}{6} - \frac{π}{4}

Least Common Multiplier of

6, 4 : 12

6, 4

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 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

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

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

6

4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

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

12

For

\frac{π}{6} :

multiply the denominator and numerator by

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

= - \frac{π 2}{12} - \frac{π 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 2 - π 3}{12}

Add similar elements:

- 2 π - 3 π = - 5 π

= \frac{- 5 π}{12}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - π + 2 πn - \frac{5 π}{12}

x \geq - π + 2 πn - \frac{5 π}{12}

x \geq - π + 2 πn - \frac{5 π}{12}

x \geq - π + 2 πn - \frac{5 π}{12}

Simplify

- π - \frac{5 π}{12} : - \frac{17 π}{12}

- π - \frac{5 π}{12}

Convert element to fraction:

π = \frac{π 12}{12}

= - \frac{π 12}{12} - \frac{5 π}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 12 - 5 π}{12}

Add similar elements:

- 12 π - 5 π = - 17 π

= \frac{- 17 π}{12}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{17 π}{12}

x \geq - \frac{17 π}{12} + 2 πn

x + \frac{π}{4} \leq arcsin (\frac{1}{2}) + 2 πn : x \leq 2 πn - \frac{π}{12}

x + \frac{π}{4} \leq arcsin (\frac{1}{2}) + 2 πn

Simplify

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

arcsin (\frac{1}{2}) + 2 πn

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} + 2 πn

x + \frac{π}{4} \leq \frac{π}{6} + 2 πn

Move

\frac{π}{4}

to the right side

x + \frac{π}{4} \leq \frac{π}{6} + 2 πn

Subtract

\frac{π}{4}

from both sides

x + \frac{π}{4} - \frac{π}{4} \leq \frac{π}{6} + 2 πn - \frac{π}{4}

Simplify

x + \frac{π}{4} - \frac{π}{4} \leq \frac{π}{6} + 2 πn - \frac{π}{4}

Simplify

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} \leq 0

= x

Simplify

\frac{π}{6} + 2 πn - \frac{π}{4} : 2 πn - \frac{π}{12}

\frac{π}{6} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn + \frac{π}{6} - \frac{π}{4}

Least Common Multiplier of

6, 4 : 12

6, 4

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 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

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

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

6

4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

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

12

For

\frac{π}{6} :

multiply the denominator and numerator by

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

For

\frac{π}{4} :

multiply the denominator and numerator by

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

= \frac{π 2}{12} - \frac{π 3}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{π 2 - π 3}{12}

Add similar elements:

2 π - 3 π = - π

= \frac{- π}{12}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= 2 πn - \frac{π}{12}

x \leq 2 πn - \frac{π}{12}

x \leq 2 πn - \frac{π}{12}

x \leq 2 πn - \frac{π}{12}

Combine the intervals

x \geq - \frac{17 π}{12} + 2 πn and x \leq 2 πn - \frac{π}{12}

Merge Overlapping Intervals

- \frac{17 π}{12} + 2 πn \leq x \leq - \frac{π}{12} + 2 πn

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