What is the general solution for cos(2θ)= 1/4 ?

The general solution for cos(2θ)= 1/4 is θ=(1.31811…)/2+pin,θ=pi-(1.31811…)/2+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2θ)= 1/4

cos (2 θ) = \frac{1}{4}

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

Solution steps

cos (2 θ) = \frac{1}{4}

Apply trig inverse properties

2 θ = arccos (\frac{1}{4}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Solve

2 θ = arccos (\frac{1}{4}) + 2 πn : θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

Solve

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn : θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn, θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

Show solutions in decimal form

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

sec (θ) = \frac{25}{7}

(3 + sin (x))^{2} = 13 - cos^{2} (x)

sec (x) = 3, \frac{3 π}{2} \leq x \leq 2 π

tan (2 x) = cot (3 x)

sin (x) = \infty

What is the general solution for cos(2θ)= 1/4 ?
The general solution for cos(2θ)= 1/4 is θ=(1.31811…)/2+pin,θ=pi-(1.31811…)/2+pin