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Popular Trigonometry >

sin(5x-30)<= (sqrt(3))/2

Solution

sin (5 x - 3 0^{\circ}) \leq \frac{3}{2}

Solution

\frac{- 4 π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n \leq x \leq \frac{π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

Interval Notation

[\frac{- 4 π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n, \frac{π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n]

Decimal

- 0.73303 \dots + \frac{2 π}{5} n \leq x \leq 0.31415 \dots + \frac{2 π}{5} n

Solution steps

sin (5 x - 3 0^{\circ}) \leq \frac{3}{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{3}{2}) + 2 πn \leq (5 x - 3 0^{\circ}) \leq arcsin (\frac{3}{2}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- π - arcsin (\frac{3}{2}) + 2 πn \leq 5 x - 3 0^{\circ} and 5 x - 3 0^{\circ} \leq arcsin (\frac{3}{2}) + 2 πn

- π - arcsin (\frac{3}{2}) + 2 πn \leq 5 x - 3 0^{\circ} : x \geq \frac{- 4 π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

- π - arcsin (\frac{3}{2}) + 2 πn \leq 5 x - 3 0^{\circ}

Switch sides

5 x - 3 0^{\circ} \geq - π - arcsin (\frac{3}{2}) + 2 πn

Simplify

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

- π - arcsin (\frac{3}{2}) + 2 πn

Use the following trivial identity:

arcsin (\frac{3}{2}) = \frac{π}{3}

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}{3} + 2 πn

5 x - 3 0^{\circ} \geq - π - \frac{π}{3} + 2 πn

Move

3 0^{\circ}

to the right side

5 x - 3 0^{\circ} \geq - π - \frac{π}{3} + 2 πn

Add

3 0^{\circ}

to both sides

5 x - 3 0^{\circ} + 3 0^{\circ} \geq - π - \frac{π}{3} + 2 πn + 3 0^{\circ}

Simplify

5 x \geq - π - \frac{π}{3} + 2 πn + 3 0^{\circ}

5 x \geq - π - \frac{π}{3} + 2 πn + 3 0^{\circ}

Divide both sides by

5

5 x \geq - π - \frac{π}{3} + 2 πn + 3 0^{\circ}

Divide both sides by

5

\frac{5 x}{5} \geq - \frac{π}{5} - \frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5}

Simplify

\frac{5 x}{5} \geq - \frac{π}{5} - \frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5}

Simplify

\frac{5 x}{5} : x

\frac{5 x}{5}

Divide the numbers:

\frac{5}{5} = 1

= x

Simplify

- \frac{π}{5} - \frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5} : - \frac{π}{5} - \frac{π}{15} + \frac{2 πn}{5} + 6^{\circ}

- \frac{π}{5} - \frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5}

Group like terms

= - \frac{π}{5} + \frac{2 πn}{5} - \frac{\frac{π}{3}}{5} + \frac{3 0 ^{\circ}}{5}

\frac{\frac{π}{3}}{5} = \frac{π}{15}

\frac{\frac{π}{3}}{5}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 5}

Multiply the numbers:

3 \cdot 5 = 15

= \frac{π}{15}

\frac{3 0 ^{\circ}}{5} = 6^{\circ}

\frac{3 0 ^{\circ}}{5}

Factor

3 0^{\circ} : 2^{\circ} \cdot 3^{\circ} \cdot 5^{\circ}

Factor

30 = 2 \cdot 3 \cdot 5

= (2 \cdot 3 \cdot 5)^{\circ}

Apply exponent rule:

(ab)^{c} = a^{c} b^{c}

= 2^{\circ} \cdot 3^{\circ} \cdot 5^{\circ}

= \frac{2 ^{\circ} \cdot 3 ^{\circ} \cdot 5 ^{\circ}}{5}

Cancel the common factor:

5^{\circ}

= 2^{\circ} \cdot 3^{\circ}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

2^{\circ} \cdot 3^{\circ} = (2 \cdot 3)^{\circ}

= (2 \cdot 3)^{\circ}

Multiply the numbers:

2 \cdot 3 = 6

= 6^{\circ}

= - \frac{π}{5} + \frac{2 πn}{5} - \frac{π}{15} + 6^{\circ}

Group like terms

= - \frac{π}{5} - \frac{π}{15} + \frac{2 πn}{5} + 6^{\circ}

x \geq - \frac{π}{5} - \frac{π}{15} + \frac{2 πn}{5} + 6^{\circ}

x \geq - \frac{π}{5} - \frac{π}{15} + \frac{2 πn}{5} + 6^{\circ}

Simplify

- \frac{π}{5} - \frac{π}{15} + 6^{\circ} : \frac{- 4 π + 15 \cdot 6 ^{\circ}}{15}

- \frac{π}{5} - \frac{π}{15} + 6^{\circ}

Convert element to fraction:

6^{\circ} = \frac{6 ^{\circ}}{1}

= - \frac{π}{5} - \frac{π}{15} + \frac{6 ^{\circ}}{1}

Least Common Multiplier of

5, 15, 1 : 15

5, 15, 1

Least Common Multiplier (LCM)

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Prime factorization of

15 : 3 \cdot 5

15

15

divides by

315 = 5 \cdot 3

= 3 \cdot 5

3, 5

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 5

Prime factorization of

1

Compute a number comprised of factors that appear in at least one of the following:

5, 15, 1

= 5 \cdot 3

Multiply the numbers:

5 \cdot 3 = 15

= 15

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

15

For

\frac{π}{5} :

multiply the denominator and numerator by

3

\frac{π}{5} = \frac{π 3}{5 \cdot 3} = \frac{π 3}{15}

For

\frac{6 ^{\circ}}{1} :

multiply the denominator and numerator by

15

\frac{6 ^{\circ}}{1} = \frac{6 ^{\circ} \cdot 15}{1 \cdot 15} = \frac{6 ^{\circ} \cdot 15}{15}

= - \frac{π 3}{15} - \frac{π}{15} + \frac{6 ^{\circ} \cdot 15}{15}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 3 - π + 6 ^{\circ} \cdot 15}{15}

Add similar elements:

- 3 π - π = - 4 π

= \frac{- 4 π + 15 \cdot 6 ^{\circ}}{15}

x \geq \frac{- 4 π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

x \geq \frac{- 4 π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

5 x - 3 0^{\circ} \leq arcsin (\frac{3}{2}) + 2 πn : x \leq \frac{π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

5 x - 3 0^{\circ} \leq arcsin (\frac{3}{2}) + 2 πn

Simplify

arcsin (\frac{3}{2}) + 2 πn : \frac{π}{3} + 2 πn

arcsin (\frac{3}{2}) + 2 πn

Use the following trivial identity:

arcsin (\frac{3}{2}) = \frac{π}{3}

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}{3} + 2 πn

5 x - 3 0^{\circ} \leq \frac{π}{3} + 2 πn

Move

3 0^{\circ}

to the right side

5 x - 3 0^{\circ} \leq \frac{π}{3} + 2 πn

Add

3 0^{\circ}

to both sides

5 x - 3 0^{\circ} + 3 0^{\circ} \leq \frac{π}{3} + 2 πn + 3 0^{\circ}

Simplify

5 x \leq \frac{π}{3} + 2 πn + 3 0^{\circ}

5 x \leq \frac{π}{3} + 2 πn + 3 0^{\circ}

Divide both sides by

5

5 x \leq \frac{π}{3} + 2 πn + 3 0^{\circ}

Divide both sides by

5

\frac{5 x}{5} \leq \frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5}

Simplify

\frac{5 x}{5} \leq \frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5}

Simplify

\frac{5 x}{5} : x

\frac{5 x}{5}

Divide the numbers:

\frac{5}{5} = 1

= x

Simplify

\frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5} : \frac{2 πn}{5} + \frac{π}{15} + 6^{\circ}

\frac{\frac{π}{3}}{5} + \frac{2 πn}{5} + \frac{3 0 ^{\circ}}{5}

Group like terms

= \frac{2 πn}{5} + \frac{\frac{π}{3}}{5} + \frac{3 0 ^{\circ}}{5}

\frac{\frac{π}{3}}{5} = \frac{π}{15}

\frac{\frac{π}{3}}{5}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 5}

Multiply the numbers:

3 \cdot 5 = 15

= \frac{π}{15}

\frac{3 0 ^{\circ}}{5} = 6^{\circ}

\frac{3 0 ^{\circ}}{5}

Factor

3 0^{\circ} : 2^{\circ} \cdot 3^{\circ} \cdot 5^{\circ}

Factor

30 = 2 \cdot 3 \cdot 5

= (2 \cdot 3 \cdot 5)^{\circ}

Apply exponent rule:

(ab)^{c} = a^{c} b^{c}

= 2^{\circ} \cdot 3^{\circ} \cdot 5^{\circ}

= \frac{2 ^{\circ} \cdot 3 ^{\circ} \cdot 5 ^{\circ}}{5}

Cancel the common factor:

5^{\circ}

= 2^{\circ} \cdot 3^{\circ}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

2^{\circ} \cdot 3^{\circ} = (2 \cdot 3)^{\circ}

= (2 \cdot 3)^{\circ}

Multiply the numbers:

2 \cdot 3 = 6

= 6^{\circ}

= \frac{2 πn}{5} + \frac{π}{15} + 6^{\circ}

x \leq \frac{2 πn}{5} + \frac{π}{15} + 6^{\circ}

x \leq \frac{2 πn}{5} + \frac{π}{15} + 6^{\circ}

Simplify

\frac{π}{15} + 6^{\circ} : \frac{π + 15 \cdot 6 ^{\circ}}{15}

\frac{π}{15} + 6^{\circ}

Convert element to fraction:

6^{\circ} = \frac{6 ^{\circ} \cdot 15}{15}

= \frac{π}{15} + \frac{6 ^{\circ} \cdot 15}{15}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 6 ^{\circ} \cdot 15}{15}

x \leq \frac{π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

x \leq \frac{π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

Combine the intervals

x \geq \frac{- 4 π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n and x \leq \frac{π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

Merge Overlapping Intervals

\frac{- 4 π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n \leq x \leq \frac{π + 15 \cdot 6 ^{\circ}}{15} + \frac{2 π}{5} n

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