What is the general solution for cos(θ)= 6/(sqrt(14)*\sqrt{9)} ?

The general solution for cos(θ)= 6/(sqrt(14)*\sqrt{9)} is θ=1.00685…+2pin,θ=2pi-1.00685…+2pin

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Popular Trigonometry >

cos(θ)= 6/(sqrt(14)*\sqrt{9)}

Solution

cos (θ) = \frac{6}{14 \cdot 9}

Solution

θ = 1.00685 \dots + 2 πn, θ = 2 π - 1.00685 \dots + 2 πn

Radians

θ = 1.00685 \dots + 6.28318 \dots n, θ = 5.27633 \dots + 6.28318 \dots n

Solution steps

cos (θ) = \frac{6}{14 9}

Simplify

\frac{6}{14 9} : \frac{2}{7}

\frac{6}{14 9}

Factor

6 : 2 \cdot 3

Factor

6 = 2 \cdot 3

Factor

14 : 27

Factor

14 = 2 \cdot 7

= 2 \cdot 7

Apply radical rule:

n ab = n a n b

= 27

Factor

9 : 3

Factor

9 = 3^{2}

= 3^{2}

Simplify

3^{2} : 3

3^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 3

= 3

= \frac{2 \cdot 3}{3 2 7}

Cancel

\frac{2 \cdot 3}{2 7 \cdot 3} : \frac{2}{7}

\frac{2 \cdot 3}{2 7 \cdot 3}

Divide the numbers:

\frac{3}{3} = 1

= \frac{2}{2 7}

Apply radical rule:

n a = a^{\frac{1}{n}}

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

= \frac{2}{2 ^{\frac{1}{2}} 7}

Apply exponent rule:

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

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

= \frac{2 ^{1 - \frac{1}{2}}}{7}

Subtract the numbers:

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

= \frac{2 ^{\frac{1}{2}}}{7}

Apply radical rule:

a^{\frac{1}{n}} = n a

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

= \frac{2}{7}

= \frac{2}{7}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{2}{7}

Apply trig inverse properties

cos (θ) = \frac{2}{7}

General solutions for

cos (θ) = \frac{2}{7}

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

θ = arccos (\frac{2}{7}) + 2 πn, θ = 2 π - arccos (\frac{2}{7}) + 2 πn

θ = arccos (\frac{2}{7}) + 2 πn, θ = 2 π - arccos (\frac{2}{7}) + 2 πn

Show solutions in decimal form

θ = 1.00685 \dots + 2 πn, θ = 2 π - 1.00685 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)= 6/(sqrt(14)*\sqrt{9)} ?
The general solution for cos(θ)= 6/(sqrt(14)*\sqrt{9)} is θ=1.00685…+2pin,θ=2pi-1.00685…+2pin