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Popular >

cos(x-45)< 1/2 ,0<= x<= 360

Solution

cos (x - 4 5^{\circ}) < \frac{1}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

\frac{π + 3 \cdot 4 5 ^{\circ}}{3} < x < \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3}

Interval Notation

(\frac{π + 34 5 ^{\circ}}{3}, \frac{5 π + 34 5 ^{\circ}}{3})

Decimal Notation

1.83259 \dots < x < 6.02138 \dots

Solution steps

cos (x - 4 5^{\circ}) < \frac{1}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

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

a < u < b

then

a < u and u < b

arccos (\frac{1}{2}) + 2 πn < x - 4 5^{\circ} and x - 4 5^{\circ} < 2 π - arccos (\frac{1}{2}) + 2 πn

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

arccos (\frac{1}{2}) + 2 πn < x - 4 5^{\circ}

Switch sides

x - 4 5^{\circ} > arccos (\frac{1}{2}) + 2 πn

Simplify

arccos (\frac{1}{2}) + 2 πn : \frac{π}{3} + 2 πn

arccos (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3} + 2 πn

x - 4 5^{\circ} > \frac{π}{3} + 2 πn

Move

4 5^{\circ}

to the right side

x - 4 5^{\circ} > \frac{π}{3} + 2 πn

Add

4 5^{\circ}

to both sides

x - 4 5^{\circ} + 4 5^{\circ} > \frac{π}{3} + 2 πn + 4 5^{\circ}

Simplify

x > \frac{π}{3} + 2 πn + 4 5^{\circ}

x > \frac{π}{3} + 2 πn + 4 5^{\circ}

Simplify

\frac{π}{3} + 4 5^{\circ} : \frac{π + 3 \cdot 4 5 ^{\circ}}{3}

\frac{π}{3} + 4 5^{\circ}

Convert element to fraction:

4 5^{\circ} = \frac{4 5 ^{\circ} \cdot 3}{3}

= \frac{π}{3} + \frac{4 5 ^{\circ} \cdot 3}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 4 5 ^{\circ} \cdot 3}{3}

x > \frac{π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn

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

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

Simplify

2 π - arccos (\frac{1}{2}) + 2 πn : 2 π - \frac{π}{3} + 2 πn

2 π - arccos (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}

= 2 π - \frac{π}{3} + 2 πn

x - 4 5^{\circ} < 2 π - \frac{π}{3} + 2 πn

Move

4 5^{\circ}

to the right side

x - 4 5^{\circ} < 2 π - \frac{π}{3} + 2 πn

Add

4 5^{\circ}

to both sides

x - 4 5^{\circ} + 4 5^{\circ} < 2 π - \frac{π}{3} + 2 πn + 4 5^{\circ}

Simplify

x < 2 π - \frac{π}{3} + 2 πn + 4 5^{\circ}

x < 2 π - \frac{π}{3} + 2 πn + 4 5^{\circ}

Simplify

2 π - \frac{π}{3} + 4 5^{\circ} : \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3}

2 π - \frac{π}{3} + 4 5^{\circ}

Convert element to fraction:

2 π = \frac{2 π 3}{3}, 4 5^{\circ} = \frac{4 5 ^{\circ} \cdot 3}{3}

= \frac{2 π 3}{3} - \frac{π}{3} + \frac{4 5 ^{\circ} \cdot 3}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 3 - π + 4 5 ^{\circ} \cdot 3}{3}

2 π 3 - π + 4 5^{\circ} \cdot 3 = 5 π + 3 \cdot 4 5^{\circ}

2 π 3 - π + 4 5^{\circ} \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6 π - π + 3 \cdot 4 5^{\circ}

Add similar elements:

6 π - π = 5 π

= 5 π + 3 \cdot 4 5^{\circ}

= \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3}

x < \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn

Combine the intervals

x > \frac{π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn and x < \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn

Merge Overlapping Intervals

\frac{π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn < x < \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn

Combine the intervals

\frac{π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn < x < \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3} + 2 πn and 0^{\circ} \leq x \leq 36 0^{\circ}

\frac{π + 3 \cdot 4 5 ^{\circ}}{3} < x < \frac{5 π + 3 \cdot 4 5 ^{\circ}}{3}

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