What is the general solution for -sqrt(3)+3=2sqrt(6)cos(θ) ?

The general solution for -sqrt(3)+3=2sqrt(6)cos(θ) is θ=1.30899…+2pin,θ=2pi-1.30899…+2pin

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Popular Trigonometry >

-sqrt(3)+3=2sqrt(6)cos(θ)

Solution

- 3 + 3 = 26 cos (θ)

Solution

θ = 1.30899 \dots + 2 πn, θ = 2 π - 1.30899 \dots + 2 πn

Degrees

θ = 7 5^{\circ} + 36 0^{\circ} n, θ = 28 5^{\circ} + 36 0^{\circ} n

Solution steps

- 3 + 3 = 26 cos (θ)

Switch sides

26 cos (θ) = - 3 + 3

Divide both sides by

26

26 cos (θ) = - 3 + 3

Divide both sides by

26

\frac{2 6 cos ( θ )}{2 6} = - \frac{3}{2 6} + \frac{3}{2 6}

Simplify

\frac{2 6 cos ( θ )}{2 6} = - \frac{3}{2 6} + \frac{3}{2 6}

Simplify

\frac{2 6 cos ( θ )}{2 6} : cos (θ)

\frac{2 6 cos ( θ )}{2 6}

Divide the numbers:

\frac{2}{2} = 1

= \frac{6 cos ( θ )}{6}

Cancel the common factor:

6

= cos (θ)

Simplify

- \frac{3}{2 6} + \frac{3}{2 6} : \frac{2 ( - 1 + 3 )}{4}

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

Apply rule

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

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

Factor

- 3 + 3 : 3 (- 1 + 3)

- 3 + 3

3 = 33

= - 3 + 33

Factor out common term

3

= 3 (- 1 + 3)

= \frac{3 ( - 1 + 3 )}{2 6}

Factor

6 : 23

Factor

6 = 2 \cdot 3

= 2 \cdot 3

Apply radical rule:

= 23

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

Cancel the common factor:

3

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

Rationalize

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

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

Multiply by the conjugate

\frac{2}{2}

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

222 = 2^{2}

222

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

Add similar elements:

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

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

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

2 \cdot \frac{1}{2}

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2^{2}

= \frac{2 ( - 1 + 3 )}{2 ^{2}}

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

2^{2} = 4

= \frac{2 ( 3 - 1 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

Apply trig inverse properties

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

General solutions for

cos (θ) = \frac{2 ( - 1 + 3 )}{4}

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

θ = arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn, θ = 2 π - arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn

θ = arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn, θ = 2 π - arccos (\frac{2 ( - 1 + 3 )}{4}) + 2 πn

Show solutions in decimal form

θ = 1.30899 \dots + 2 πn, θ = 2 π - 1.30899 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -sqrt(3)+3=2sqrt(6)cos(θ) ?
The general solution for -sqrt(3)+3=2sqrt(6)cos(θ) is θ=1.30899…+2pin,θ=2pi-1.30899…+2pin