What is the general solution for cos(θ)=(-3)/(sqrt(13)) ?

The general solution for cos(θ)=(-3)/(sqrt(13)) is θ=2.55359…+2pin,θ=-2.55359…+2pin

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cos(θ)=(-3)/(sqrt(13))

cos (θ) = \frac{- 3}{13}

θ = 2.55359 \dots + 2 πn, θ = - 2.55359 \dots + 2 πn

Radians

θ = 2.55359 \dots + 6.28318 \dots n, θ = - 2.55359 \dots + 6.28318 \dots n

Solution steps

cos (θ) = \frac{- 3}{13}

Simplify

\frac{- 3}{13} : - \frac{3 13}{13}

\frac{- 3}{13}

Apply the fraction rule:

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

= - \frac{3}{13}

Rationalize

- \frac{3}{13} : - \frac{3 13}{13}

- \frac{3}{13}

Multiply by the conjugate

\frac{13}{13}

= - \frac{3 13}{13 13}

1313 = 13

1313

Apply radical rule:

a a = a

1313 = 13

= 13

= - \frac{3 13}{13}

= - \frac{3 13}{13}

cos (θ) = - \frac{3 13}{13}

Apply trig inverse properties

cos (θ) = - \frac{3 13}{13}

General solutions for

cos (θ) = - \frac{3 13}{13}

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

θ = arccos (- \frac{3 13}{13}) + 2 πn, θ = - arccos (- \frac{3 13}{13}) + 2 πn

θ = arccos (- \frac{3 13}{13}) + 2 πn, θ = - arccos (- \frac{3 13}{13}) + 2 πn

Show solutions in decimal form

θ = 2.55359 \dots + 2 πn, θ = - 2.55359 \dots + 2 πn

cos (x) = 0.84

cos (2 x) = \frac{( 4 k - 5 )}{10}

cos^{2} (2 x) - sin^{2} (2 x) = - 1

cos (x + \frac{π}{4}) = 1

arctan (x) = - 180

What is the general solution for cos(θ)=(-3)/(sqrt(13)) ?
The general solution for cos(θ)=(-3)/(sqrt(13)) is θ=2.55359…+2pin,θ=-2.55359…+2pin