What is the general solution for 5cos(2θ)=2 ?

The general solution for 5cos(2θ)=2 is θ=(1.15927…)/2+pin,θ=pi-(1.15927…)/2+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5cos(2θ)=2

5 cos (2 θ) = 2

θ = \frac{1.15927 \dots}{2} + πn, θ = π - \frac{1.15927 \dots}{2} + πn

Solution steps

5 cos (2 θ) = 2

Divide both sides by

5

cos (2 θ) = \frac{2}{5}

Apply trig inverse properties

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

Solve

2 θ = arccos (\frac{2}{5}) + 2 πn : θ = \frac{arccos ( \frac{2}{5} )}{2} + πn

Solve

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

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

Show solutions in decimal form

θ = \frac{1.15927 \dots}{2} + πn, θ = π - \frac{1.15927 \dots}{2} + πn

2 cos^{2} (θ) = cos (θ) + 1

tan (θ) = \frac{100}{101}

cos (\frac{π}{2} + x) + sin (\frac{π}{2} + x) = 0

10 tan (b) + 12 = tan (b) + 6

arctan (x) + arctan (1 - x) = arctan (\frac{4}{3})

What is the general solution for 5cos(2θ)=2 ?
The general solution for 5cos(2θ)=2 is θ=(1.15927…)/2+pin,θ=pi-(1.15927…)/2+pin