What is the general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) ?

The general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) is θ=1.45945…+2pin,θ=2pi-1.45945…+2pin

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cos(θ)= 1/(sqrt(1^2+(4)^2+8^2))

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

Solution steps

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

Simplify

\frac{1}{1 ^{2} + 4 ^{2} + 8 ^{2}} : \frac{1}{9}

Apply trig inverse properties

θ = arccos (\frac{1}{9}) + 2 πn, θ = 2 π - arccos (\frac{1}{9}) + 2 πn

Show solutions in decimal form

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

2 + 3 cot^{2} (x) - 3 csc (x) = 5

- 2 = - 2 + tan (θ)

cos (A) = 0.65

sin (3 x - \frac{π}{4}) = \frac{2}{2}

sin (2 θ) = \frac{- 3}{4}, \frac{3 π}{2} \leq 2 θ \leq 2 π

What is the general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) ?
The general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) is θ=1.45945…+2pin,θ=2pi-1.45945…+2pin