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Popular Trigonometry >

3/2 cos(x)-2>= cos(x)-7/4

Solution

\frac{3}{2} cos (x) - 2 \geq cos (x) - \frac{7}{4}

Solution

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

Interval Notation

[- \frac{π}{3} + 2 πn, \frac{π}{3} + 2 πn]

Decimal

- 1.04719 \dots + 2 πn \leq x \leq 1.04719 \dots + 2 πn

Solution steps

\frac{3}{2} cos (x) - 2 \geq cos (x) - \frac{7}{4}

Let:

u = cos (x)

\frac{3}{2} u - 2 \geq u - \frac{7}{4}

\frac{3}{2} u - 2 \geq u - \frac{7}{4} : u \geq \frac{1}{2}

\frac{3}{2} u - 2 \geq u - \frac{7}{4}

Move

2

to the right side

\frac{3}{2} u - 2 \geq u - \frac{7}{4}

Add

2

to both sides

\frac{3}{2} u - 2 + 2 \geq u - \frac{7}{4} + 2

Simplify

\frac{3}{2} u - 2 + 2 \geq u - \frac{7}{4} + 2

Simplify

\frac{3}{2} u - 2 + 2 : \frac{3}{2} u

\frac{3}{2} u - 2 + 2

Add similar elements:

- 2 + 2 \geq 0

= \frac{3}{2} u

Simplify

u - \frac{7}{4} + 2 : u + \frac{1}{4}

u - \frac{7}{4} + 2

Combine the fractions

2 - \frac{7}{4} : \frac{1}{4}

2 - \frac{7}{4}

Convert element to fraction:

2 = \frac{2 \cdot 4}{4}

= \frac{2 \cdot 4}{4} - \frac{7}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{2 \cdot 4 - 7}{4}

2 \cdot 4 - 7 = 1

2 \cdot 4 - 7

Multiply the numbers:

2 \cdot 4 = 8

= 8 - 7

Subtract the numbers:

8 - 7 = 1

= 1

= \frac{1}{4}

= u + \frac{1}{4}

\frac{3}{2} u \geq u + \frac{1}{4}

\frac{3}{2} u \geq u + \frac{1}{4}

\frac{3}{2} u \geq u + \frac{1}{4}

Move

u

to the left side

\frac{3}{2} u \geq u + \frac{1}{4}

Subtract

u

from both sides

\frac{3}{2} u - u \geq u + \frac{1}{4} - u

Simplify

\frac{3}{2} u - u \geq u + \frac{1}{4} - u

Simplify

\frac{3}{2} u - u : \frac{1}{2} u

\frac{3}{2} u - u

Factor out common term

u

= u (\frac{3}{2} - 1)

\frac{3}{2} - 1 = \frac{1}{2}

\frac{3}{2} - 1

Convert element to fraction:

1 = \frac{1 \cdot 2}{2}

= \frac{3}{2} - \frac{1 \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{3 - 1 \cdot 2}{2}

3 - 1 \cdot 2 = 1

3 - 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 3 - 2

Subtract the numbers:

3 - 2 = 1

= 1

= \frac{1}{2}

= \frac{1}{2} u

Simplify

u + \frac{1}{4} - u : \frac{1}{4}

u + \frac{1}{4} - u

Add similar elements:

u - u \geq 0

= \frac{1}{4}

\frac{1}{2} u \geq \frac{1}{4}

\frac{1}{2} u \geq \frac{1}{4}

\frac{1}{2} u \geq \frac{1}{4}

Multiply both sides by

2

\frac{1}{2} u \geq \frac{1}{4}

Multiply both sides by

2

2 \cdot \frac{1}{2} u \geq \frac{1 \cdot 2}{4}

Simplify

2 \cdot \frac{1}{2} u \geq \frac{1 \cdot 2}{4}

Simplify

2 \cdot \frac{1}{2} u : u

2 \cdot \frac{1}{2} u

Multiply fractions:

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

= \frac{1 \cdot 2}{2} u

Cancel the common factor:

2

= u \cdot 1

Multiply:

u \cdot 1 = u

= u

Simplify

\frac{1 \cdot 2}{4} : \frac{1}{2}

\frac{1 \cdot 2}{4}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u \geq \frac{1}{2}

u \geq \frac{1}{2}

u \geq \frac{1}{2}

u \geq \frac{1}{2}

Substitute back

u = cos (x)

cos (x) \geq \frac{1}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (\frac{1}{2}) + 2 πn \leq x \leq arccos (\frac{1}{2}) + 2 πn

Simplify

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

- arccos (\frac{1}{2})

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}

Simplify

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

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}

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

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