What is the general solution for cos(θ)= 1/(sqrt(10)) ?

The general solution for cos(θ)= 1/(sqrt(10)) is θ=1.24904…+2pin,θ=2pi-1.24904…+2pin

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cos(θ)= 1/(sqrt(10))

cos (θ) = \frac{1}{10}

θ = 1.24904 \dots + 2 πn, θ = 2 π - 1.24904 \dots + 2 πn

Radians

θ = 1.24904 \dots + 6.28318 \dots n, θ = 5.03413 \dots + 6.28318 \dots n

Solution steps

cos (θ) = \frac{1}{10}

Simplify

\frac{1}{10} : \frac{10}{10}

\frac{1}{10}

Multiply by the conjugate

\frac{10}{10}

= \frac{1 \cdot 10}{10 10}

1 \cdot 10 = 10

1010 = 10

1010

Apply radical rule:

a a = a

1010 = 10

= 10

= \frac{10}{10}

cos (θ) = \frac{10}{10}

Apply trig inverse properties

cos (θ) = \frac{10}{10}

General solutions for

cos (θ) = \frac{10}{10}

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

θ = arccos (\frac{10}{10}) + 2 πn, θ = 2 π - arccos (\frac{10}{10}) + 2 πn

θ = arccos (\frac{10}{10}) + 2 πn, θ = 2 π - arccos (\frac{10}{10}) + 2 πn

Show solutions in decimal form

θ = 1.24904 \dots + 2 πn, θ = 2 π - 1.24904 \dots + 2 πn

2 = 2 sin (x)

cos^{2} (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

cot (θ) - 3 = 0, 0 \leq θ \leq 2 π

2 = sec (θ)

sin (x) = \frac{1}{8}, sin (2 x)

What is the general solution for cos(θ)= 1/(sqrt(10)) ?
The general solution for cos(θ)= 1/(sqrt(10)) is θ=1.24904…+2pin,θ=2pi-1.24904…+2pin