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Popular >

-2cos(2x-30)<0

Solution

- 2 cos (2 x - 3 0^{\circ}) < 0

Solution

\frac{- π + 4 \cdot 1 5 ^{\circ}}{4} + πn < x < \frac{π + 4 \cdot 1 5 ^{\circ}}{4} + πn

Interval Notation

(\frac{- π + 4 \cdot 1 5 ^{\circ}}{4} + πn, \frac{π + 4 \cdot 1 5 ^{\circ}}{4} + πn)

Decimal Notation

- 0.52359 \dots + πn < x < 1.04719 \dots + πn

Solution steps

- 2 cos (2 x - 3 0^{\circ}) < 0

Multiply both sides by

- 1

- 2 cos (2 x - 3 0^{\circ}) < 0

Multiply both sides by -1 (reverse the inequality)

(- 2 cos (2 x - 3 0^{\circ})) (- 1) > 0 \cdot (- 1)

Simplify

2 cos (2 x - 3 0^{\circ}) > 0

2 cos (2 x - 3 0^{\circ}) > 0

Divide both sides by

2

2 cos (2 x - 3 0^{\circ}) > 0

Divide both sides by

2

\frac{2 cos ( 2 x - 3 0 ^{\circ} )}{2} > \frac{0}{2}

Simplify

cos (2 x - 3 0^{\circ}) > 0

cos (2 x - 3 0^{\circ}) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < (2 x - 3 0^{\circ}) < arccos (0) + 2 πn

a < u < b

then

a < u and u < b

- arccos (0) + 2 πn < 2 x - 3 0^{\circ} and 2 x - 3 0^{\circ} < arccos (0) + 2 πn

- arccos (0) + 2 πn < 2 x - 3 0^{\circ} : x > \frac{- π + 4 \cdot 1 5 ^{\circ}}{4} + πn

- arccos (0) + 2 πn < 2 x - 3 0^{\circ}

Switch sides

2 x - 3 0^{\circ} > - arccos (0) + 2 πn

Simplify

- arccos (0) + 2 πn : - \frac{π}{2} + 2 πn

- arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2} + 2 πn

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

Move

3 0^{\circ}

to the right side

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

Add

3 0^{\circ}

to both sides

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

Simplify

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

- \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{3 0 ^{\circ}}{2} : πn - \frac{π}{4} + 1 5^{\circ}

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

Group like terms

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{3 0 ^{\circ}}{2} = 1 5^{\circ}

\frac{3 0 ^{\circ}}{2}

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}}{2}

Cancel the common factor:

2^{\circ}

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

Apply exponent rule:

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

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

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

Multiply the numbers:

3 \cdot 5 = 15

= 1 5^{\circ}

= πn - \frac{π}{4} + 1 5^{\circ}

x > πn - \frac{π}{4} + 1 5^{\circ}

x > πn - \frac{π}{4} + 1 5^{\circ}

Simplify

- \frac{π}{4} + 1 5^{\circ} : \frac{- π + 4 \cdot 1 5 ^{\circ}}{4}

- \frac{π}{4} + 1 5^{\circ}

Convert element to fraction:

1 5^{\circ} = \frac{1 5 ^{\circ} \cdot 4}{4}

= - \frac{π}{4} + \frac{1 5 ^{\circ} \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{- π + 1 5 ^{\circ} \cdot 4}{4}

x > \frac{- π + 4 \cdot 1 5 ^{\circ}}{4} + πn

x > \frac{- π + 4 \cdot 1 5 ^{\circ}}{4} + πn

2 x - 3 0^{\circ} < arccos (0) + 2 πn : x < \frac{π + 4 \cdot 1 5 ^{\circ}}{4} + πn

2 x - 3 0^{\circ} < arccos (0) + 2 πn

Simplify

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2} + 2 πn

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

Move

3 0^{\circ}

to the right side

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

Add

3 0^{\circ}

to both sides

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

Simplify

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} + \frac{3 0 ^{\circ}}{2} : πn + \frac{π}{4} + 1 5^{\circ}

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

Group like terms

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{3 0 ^{\circ}}{2} = 1 5^{\circ}

\frac{3 0 ^{\circ}}{2}

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}}{2}

Cancel the common factor:

2^{\circ}

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

Apply exponent rule:

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

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

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

Multiply the numbers:

3 \cdot 5 = 15

= 1 5^{\circ}

= πn + \frac{π}{4} + 1 5^{\circ}

x < πn + \frac{π}{4} + 1 5^{\circ}

x < πn + \frac{π}{4} + 1 5^{\circ}

Simplify

\frac{π}{4} + 1 5^{\circ} : \frac{π + 4 \cdot 1 5 ^{\circ}}{4}

\frac{π}{4} + 1 5^{\circ}

Convert element to fraction:

1 5^{\circ} = \frac{1 5 ^{\circ} \cdot 4}{4}

= \frac{π}{4} + \frac{1 5 ^{\circ} \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 1 5 ^{\circ} \cdot 4}{4}

x < \frac{π + 4 \cdot 1 5 ^{\circ}}{4} + πn

x < \frac{π + 4 \cdot 1 5 ^{\circ}}{4} + πn

Combine the intervals

x > \frac{- π + 4 \cdot 1 5 ^{\circ}}{4} + πn and x < \frac{π + 4 \cdot 1 5 ^{\circ}}{4} + πn

Merge Overlapping Intervals

\frac{- π + 4 \cdot 1 5 ^{\circ}}{4} + πn < x < \frac{π + 4 \cdot 1 5 ^{\circ}}{4} + πn

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