What is the general solution for cos(θ)=-0.8,sin(pi^2-θ) ?

The general solution for cos(θ)=-0.8,sin(pi^2-θ) is No Solution for θ\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(θ)=-0.8,sin(pi^2-θ)

cos (θ) = - 0.8, sin (π^{2} - θ)

No Solution for θ \in R

Solution steps

cos (θ) = - 0.8, sin (π^{2} - θ)

Apply trig inverse properties

cos (θ) = - 0.8

General solutions for

cos (θ) = - 0.8

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

θ = arccos (- 0.8) + 2 πn, θ = - arccos (- 0.8) + 2 πn

θ = arccos (- 0.8) + 2 πn, θ = - arccos (- 0.8) + 2 πn

Solutions for the range

sin (π^{2} - θ)

No Solution for θ \in R

0 = a (1 + cos (2 θ))

4 sin (2 x) = 3 cos (x)

sin (b) = (\frac{250 km}{590 km}) sin (2 3^{\circ}), 8 2^{\circ}

sin (x) = \frac{30}{60}

4 cos (z) + 5 = 0

What is the general solution for cos(θ)=-0.8,sin(pi^2-θ) ?
The general solution for cos(θ)=-0.8,sin(pi^2-θ) is No Solution for θ\in\mathbb{R}