What is the general solution for sqrt((-(\sqrt{2))/3)^2+1}=sec(θ) ?

The general solution for sqrt((-(\sqrt{2))/3)^2+1}=sec(θ) is θ=0.44051…+2pin,θ=2pi-0.44051…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt((-(\sqrt{2))/3)^2+1}=sec(θ)

(- \frac{2}{3})^{2} + 1 = sec (θ)

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

Solution steps

(- \frac{2}{3})^{2} + 1 = sec (θ)

Switch sides

sec (θ) = (- \frac{2}{3})^{2} + 1

Simplify

(- \frac{2}{3})^{2} + 1 : \frac{11}{3}

Apply trig inverse properties

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

Show solutions in decimal form

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

1 = sech^{2} (x)

tan (θ) = 0.86667

sin (x) \cdot cos^{2} (x) = 1

t, f w = 2 + cos (10 π t)

tan (θ) - sec (θ) = 3

What is the general solution for sqrt((-(\sqrt{2))/3)^2+1}=sec(θ) ?
The general solution for sqrt((-(\sqrt{2))/3)^2+1}=sec(θ) is θ=0.44051…+2pin,θ=2pi-0.44051…+2pin