What is the general solution for cos(θ)= 3/(sqrt(11)) ?

The general solution for cos(θ)= 3/(sqrt(11)) is θ=0.44051…+2pin,θ=2pi-0.44051…+2pin

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cos(θ)= 3/(sqrt(11))

cos (θ) = \frac{3}{11}

θ = 0.44051 \dots + 2 πn, θ = 2 π - 0.44051 \dots + 2 πn

Radians

θ = 0.44051 \dots + 6.28318 \dots n, θ = 5.84267 \dots + 6.28318 \dots n

Solution steps

cos (θ) = \frac{3}{11}

Simplify

\frac{3}{11} : \frac{3 11}{11}

\frac{3}{11}

Multiply by the conjugate

\frac{11}{11}

= \frac{3 11}{11 11}

1111 = 11

1111

Apply radical rule:

a a = a

1111 = 11

= 11

= \frac{3 11}{11}

cos (θ) = \frac{3 11}{11}

Apply trig inverse properties

cos (θ) = \frac{3 11}{11}

General solutions for

cos (θ) = \frac{3 11}{11}

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

θ = arccos (\frac{3 11}{11}) + 2 πn, θ = 2 π - arccos (\frac{3 11}{11}) + 2 πn

θ = arccos (\frac{3 11}{11}) + 2 πn, θ = 2 π - arccos (\frac{3 11}{11}) + 2 πn

Show solutions in decimal form

θ = 0.44051 \dots + 2 πn, θ = 2 π - 0.44051 \dots + 2 πn

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\frac{sec ( x )}{cos ( x )} - 1 = 0

- 9 (sin (x))^{2} + 2 sin (x) + 2 = 0

cos^{2} (x) - 2 cos (x) + 1 = 0, 0^{\circ} 36 0^{\circ}

sin^{4} (x) = sin^{2} (x), 0 \leq x < 2 π

What is the general solution for cos(θ)= 3/(sqrt(11)) ?
The general solution for cos(θ)= 3/(sqrt(11)) is θ=0.44051…+2pin,θ=2pi-0.44051…+2pin