What is the general solution for cos(θ)= 3/(sqrt(45)) ?

The general solution for cos(θ)= 3/(sqrt(45)) is θ=1.10714…+2pin,θ=2pi-1.10714…+2pin

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cos(θ)= 3/(sqrt(45))

cos (θ) = \frac{3}{45}

θ = 1.10714 \dots + 2 πn, θ = 2 π - 1.10714 \dots + 2 πn

Radians

θ = 1.10714 \dots + 6.28318 \dots n, θ = 5.17603 \dots + 6.28318 \dots n

Solution steps

cos (θ) = \frac{3}{45}

Simplify

\frac{3}{45} : \frac{5}{5}

\frac{3}{45}

Factor

45 : 35

Factor

45 = 3^{2} \cdot 5

= 3^{2} \cdot 5

Apply radical rule:

= 5 3^{2}

Simplify

3^{2} : 3

3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

= 35

= \frac{3}{3 5}

Divide the numbers:

\frac{3}{3} = 1

= \frac{1}{5}

Rationalize

\frac{1}{5} : \frac{5}{5}

\frac{1}{5}

Multiply by the conjugate

\frac{5}{5}

= \frac{1 \cdot 5}{5 5}

1 \cdot 5 = 5

55 = 5

55

Apply radical rule:

a a = a

55 = 5

= 5

= \frac{5}{5}

= \frac{5}{5}

cos (θ) = \frac{5}{5}

Apply trig inverse properties

cos (θ) = \frac{5}{5}

General solutions for

cos (θ) = \frac{5}{5}

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

θ = arccos (\frac{5}{5}) + 2 πn, θ = 2 π - arccos (\frac{5}{5}) + 2 πn

θ = arccos (\frac{5}{5}) + 2 πn, θ = 2 π - arccos (\frac{5}{5}) + 2 πn

Show solutions in decimal form

θ = 1.10714 \dots + 2 πn, θ = 2 π - 1.10714 \dots + 2 πn

cos^{2} (x) = \frac{9}{16}

sin (4 x) = 0.5

3 cos^{2} (x) + 6 cos (x) = - 3, 0 \leq x \leq 2 π

tan (θ) = 7, sec^{2} (θ)

4 cos (x - 4 5^{\circ}) + 7 = 10, 0^{\circ} \leq x \leq 36 0^{\circ}

What is the general solution for cos(θ)= 3/(sqrt(45)) ?
The general solution for cos(θ)= 3/(sqrt(45)) is θ=1.10714…+2pin,θ=2pi-1.10714…+2pin