What is the general solution for 3cos(θ)=2,1 ?

The general solution for 3cos(θ)=2,1 is No Solution for θ\in\mathbb{R}

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3cos(θ)=2,1

3 cos (θ) = 2, 1

No Solution for θ \in R

Solution steps

3 cos (θ) = 2, 1

Divide both sides by

3

3 cos (θ) = 2

Divide both sides by

3

\frac{3 cos ( θ )}{3} = \frac{2}{3}

Simplify

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{3}

Apply trig inverse properties

cos (θ) = \frac{2}{3}

General solutions for

cos (θ) = \frac{2}{3}

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

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

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

Solutions for the range

1

No Solution for θ \in R

3 csc (x) - 7 = 0

tan (2 θ) = 0.6583

sin (x) = \frac{8}{12.8}

3 tan^{2} (x) - 2 = 5 sec^{2} (x) - 9

x, z = arcsin^{3} (3 x + 2 y)

What is the general solution for 3cos(θ)=2,1 ?
The general solution for 3cos(θ)=2,1 is No Solution for θ\in\mathbb{R}