What is the general solution for sqrt(3)cos(3x)+1=0 ?

The general solution for sqrt(3)cos(3x)+1=0 is x=(2.18627…)/3+(2pin)/3 ,x=-(2.18627…)/3+(2pin)/3

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Popular >

sqrt(3)cos(3x)+1=0

Solution

3 cos (3 x) + 1 = 0

Solution

x = \frac{2.18627 \dots}{3} + \frac{2 πn}{3}, x = - \frac{2.18627 \dots}{3} + \frac{2 πn}{3}

Degrees

x = 41.75479 \dots^{\circ} + 12 0^{\circ} n, x = - 41.75479 \dots^{\circ} + 12 0^{\circ} n

Solution steps

3 cos (3 x) + 1 = 0

Move

1

to the right side

3 cos (3 x) + 1 = 0

Subtract

1

from both sides

3 cos (3 x) + 1 - 1 = 0 - 1

Simplify

3 cos (3 x) = - 1

3 cos (3 x) = - 1

Divide both sides by

3

3 cos (3 x) = - 1

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 cos ( 3 x )}{3} : cos (3 x)

\frac{3 cos ( 3 x )}{3}

Cancel the common factor:

3

= cos (3 x)

Simplify

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

\frac{- 1}{3}

Apply the fraction rule:

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

= - \frac{1}{3}

Rationalize

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

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

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

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

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

3 x = arccos (- \frac{3}{3}) + 2 πn, 3 x = - arccos (- \frac{3}{3}) + 2 πn

3 x = arccos (- \frac{3}{3}) + 2 πn, 3 x = - arccos (- \frac{3}{3}) + 2 πn

Solve

3 x = arccos (- \frac{3}{3}) + 2 πn : x = \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

3 x = arccos (- \frac{3}{3}) + 2 πn

Simplify

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

arccos (- \frac{3}{3}) + 2 πn

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= \frac{1}{3}

= arccos (- \frac{1}{3}) + 2 πn

3 x = arccos (- \frac{1}{3}) + 2 πn

Divide both sides by

3

3 x = arccos (- \frac{1}{3}) + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{arccos ( - \frac{1}{3} )}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{arccos ( - \frac{1}{3} )}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

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

arccos (- \frac{1}{3}) = arccos (- \frac{3}{3})

arccos (- \frac{1}{3})

= arccos (- \frac{3}{3})

= \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

x = \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

x = \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

x = \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

Solve

3 x = - arccos (- \frac{3}{3}) + 2 πn : x = - \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

3 x = - arccos (- \frac{3}{3}) + 2 πn

Simplify

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

- arccos (- \frac{3}{3}) + 2 πn

\frac{3}{3} = \frac{1}{3}

\frac{3}{3}

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= \frac{1}{3}

= - arccos (- \frac{1}{3}) + 2 πn

3 x = - arccos (- \frac{1}{3}) + 2 πn

Divide both sides by

3

3 x = - arccos (- \frac{1}{3}) + 2 πn

Divide both sides by

3

\frac{3 x}{3} = - \frac{arccos ( - \frac{1}{3} )}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = - \frac{arccos ( - \frac{1}{3} )}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

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

arccos (- \frac{1}{3}) = arccos (- \frac{3}{3})

arccos (- \frac{1}{3})

= arccos (- \frac{3}{3})

= - \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

x = - \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

x = - \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

x = - \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

x = \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}, x = - \frac{arccos ( - \frac{3}{3} )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

x = \frac{2.18627 \dots}{3} + \frac{2 πn}{3}, x = - \frac{2.18627 \dots}{3} + \frac{2 πn}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)cos(3x)+1=0 ?
The general solution for sqrt(3)cos(3x)+1=0 is x=(2.18627…)/3+(2pin)/3 ,x=-(2.18627…)/3+(2pin)/3