What is the general solution for cos(θ)=-4/5 ,sin(θ) ?

The general solution for cos(θ)=-4/5 ,sin(θ) is No Solution for θ\in\mathbb{R}

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cos(θ)=-4/5 ,sin(θ)

cos (θ) = - \frac{4}{5}, sin (θ)

No Solution for θ \in R

Solution steps

cos (θ) = - \frac{4}{5}, sin (θ)

Apply trig inverse properties

cos (θ) = - \frac{4}{5}

General solutions for

cos (θ) = - \frac{4}{5}

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

θ = arccos (- \frac{4}{5}) + 2 πn, θ = - arccos (- \frac{4}{5}) + 2 πn

θ = arccos (- \frac{4}{5}) + 2 πn, θ = - arccos (- \frac{4}{5}) + 2 πn

Solutions for the range

sin (θ)

No Solution for θ \in R

5 sin (4 x) = 4 cos (2 x)

arccosh (x) = 1

θ, sin (3 θ) = \frac{1}{2}

0 = - 0.5 sin (\frac{x}{5})

\frac{408.4}{sin ( 13 0 ^{\circ} )} = \frac{200}{sin ( x )}

What is the general solution for cos(θ)=-4/5 ,sin(θ) ?
The general solution for cos(θ)=-4/5 ,sin(θ) is No Solution for θ\in\mathbb{R}