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Popular Trigonometry >

sqrt(3)cos(x)-sin(x)>= 1

Solution

3 cos (x) - sin (x) \geq 1

Solution

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

Interval Notation

[- \frac{π}{2} + 2 πn, \frac{π}{6} + 2 πn]

Decimal

- 1.57079 \dots + 2 πn \leq x \leq 0.52359 \dots + 2 πn

Solution steps

3 cos (x) - sin (x) \geq 1

Rewrite using trig identities

Divide both sides by

2

\frac{3 cos ( x ) - sin ( x )}{2} \geq \frac{1}{2}

Expand

\frac{3 cos ( x ) - sin ( x )}{2} : \frac{3}{2} cos (x) - \frac{1}{2} sin (x)

\frac{3 cos ( x ) - sin ( x )}{2}

Apply the fraction rule:

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

\frac{3 cos ( x ) - sin ( x )}{2} = \frac{3 cos ( x )}{2} - \frac{sin ( x )}{2}

= \frac{3 cos ( x )}{2} - \frac{sin ( x )}{2}

= \frac{3}{2} cos (x) - \frac{1}{2} sin (x)

\frac{3}{2} cos (x) - \frac{1}{2} sin (x) \geq \frac{1}{2}

\frac{3}{2} = sin (\frac{π}{3})

sin (\frac{π}{3}) cos (x) - \frac{1}{2} sin (x) \geq \frac{1}{2}

\frac{1}{2} = cos (\frac{π}{3})

sin (\frac{π}{3}) cos (x) - cos (\frac{π}{3}) sin (x) \geq \frac{1}{2}

Use the following identity:

- cos (s) sin (t) + cos (t) sin (s) = sin (s - t)

sin (\frac{π}{3} - x) \geq \frac{1}{2}

Factor out

- 1

from

\frac{π}{3} - x : - (- \frac{π}{3} + x)

sin (- (- \frac{π}{3} + x)) \geq \frac{1}{2}

Use the following identity:

sin (- x) = - sin (x)

- sin (- \frac{π}{3} + x) \geq \frac{1}{2}

Multiply both sides by

- 1

- sin (- \frac{π}{3} + x) \geq \frac{1}{2}

Multiply both sides by -1 (reverse the inequality)

(- sin (- \frac{π}{3} + x)) (- 1) \leq \frac{1 \cdot ( - 1 )}{2}

Simplify

sin (- \frac{π}{3} + x) \leq - \frac{1}{2}

sin (- \frac{π}{3} + x) \leq - \frac{1}{2}

sin (- \frac{π}{3} + x) \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 (- \frac{π}{3} + x) \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 - \frac{π}{3} + x and - \frac{π}{3} + x \leq arcsin (- \frac{1}{2}) + 2 πn

- π - arcsin (- \frac{1}{2}) + 2 πn \leq - \frac{π}{3} + x : x \geq - \frac{π}{2} + 2 πn

- π - arcsin (- \frac{1}{2}) + 2 πn \leq - \frac{π}{3} + x

Switch sides

- \frac{π}{3} + x \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

arcsin (- \frac{1}{2}) = - \frac{π}{6}

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

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

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}

= - \frac{π}{6}

= - π - (- \frac{π}{6}) + 2 πn

Apply rule

- (- a) = a

= - π + \frac{π}{6} + 2 πn

- \frac{π}{3} + x \geq - π + \frac{π}{6} + 2 πn

Move

\frac{π}{3}

to the right side

- \frac{π}{3} + x \geq - π + \frac{π}{6} + 2 πn

Add

\frac{π}{3}

to both sides

- \frac{π}{3} + x + \frac{π}{3} \geq - π + \frac{π}{6} + 2 πn + \frac{π}{3}

Simplify

- \frac{π}{3} + x + \frac{π}{3} \geq - π + \frac{π}{6} + 2 πn + \frac{π}{3}

Simplify

- \frac{π}{3} + x + \frac{π}{3} : x

- \frac{π}{3} + x + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} \geq 0

= x

Simplify

- π + \frac{π}{6} + 2 πn + \frac{π}{3} : - π + 2 πn + \frac{π}{2}

- π + \frac{π}{6} + 2 πn + \frac{π}{3}

Group like terms

= - π + 2 πn + \frac{π}{6} + \frac{π}{3}

Least Common Multiplier of

6, 3 : 6

6, 3

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

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

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

6

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

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

6

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= \frac{π}{6} + \frac{π 2}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π + π 2}{6}

Add similar elements:

π + 2 π = 3 π

= \frac{3 π}{6}

Cancel the common factor:

3

= - π + 2 πn + \frac{π}{2}

x \geq - π + 2 πn + \frac{π}{2}

x \geq - π + 2 πn + \frac{π}{2}

x \geq - π + 2 πn + \frac{π}{2}

Simplify

- π + \frac{π}{2} : - \frac{π}{2}

- π + \frac{π}{2}

Convert element to fraction:

π = \frac{π 2}{2}

= - \frac{π 2}{2} + \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 2 + π}{2}

Add similar elements:

- 2 π + π = - π

= \frac{- π}{2}

Apply the fraction rule:

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

= - \frac{π}{2}

x \geq - \frac{π}{2} + 2 πn

- \frac{π}{3} + x \leq arcsin (- \frac{1}{2}) + 2 πn : x \leq 2 πn + \frac{π}{6}

- \frac{π}{3} + x \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

arcsin (- \frac{1}{2}) = - \frac{π}{6}

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

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

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}

= - \frac{π}{6}

= - \frac{π}{6} + 2 πn

- \frac{π}{3} + x \leq - \frac{π}{6} + 2 πn

Move

\frac{π}{3}

to the right side

- \frac{π}{3} + x \leq - \frac{π}{6} + 2 πn

Add

\frac{π}{3}

to both sides

- \frac{π}{3} + x + \frac{π}{3} \leq - \frac{π}{6} + 2 πn + \frac{π}{3}

Simplify

- \frac{π}{3} + x + \frac{π}{3} \leq - \frac{π}{6} + 2 πn + \frac{π}{3}

Simplify

- \frac{π}{3} + x + \frac{π}{3} : x

- \frac{π}{3} + x + \frac{π}{3}

Add similar elements:

- \frac{π}{3} + \frac{π}{3} \leq 0

= x

Simplify

- \frac{π}{6} + 2 πn + \frac{π}{3} : 2 πn + \frac{π}{6}

- \frac{π}{6} + 2 πn + \frac{π}{3}

Group like terms

= 2 πn - \frac{π}{6} + \frac{π}{3}

Least Common Multiplier of

6, 3 : 6

6, 3

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

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

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

6

3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

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

6

For

\frac{π}{3} :

multiply the denominator and numerator by

2

\frac{π}{3} = \frac{π 2}{3 \cdot 2} = \frac{π 2}{6}

= - \frac{π}{6} + \frac{π 2}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{- π + π 2}{6}

Add similar elements:

- π + 2 π = π

= 2 πn + \frac{π}{6}

x \leq 2 πn + \frac{π}{6}

x \leq 2 πn + \frac{π}{6}

x \leq 2 πn + \frac{π}{6}

Combine the intervals

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

Merge Overlapping Intervals

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

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