What is the general solution for cos(3θ)=-1,0<= θ<2pi ?

The general solution for cos(3θ)=-1,0<= θ<2pi is θ= pi/3 ,θ=pi,θ=(5pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(3θ)=-1,0<= θ<2pi

cos (3 θ) = - 1, 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = π, θ = \frac{5 π}{3}

Solution steps

cos (3 θ) = - 1, 0 \leq θ < 2 π

General solutions for

cos (3 θ) = - 1

3 θ = π + 2 πn

Solve

3 θ = π + 2 πn : θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{π}{3} + \frac{2 πn}{3}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = π, θ = \frac{5 π}{3}

2 cos (x) + 3 = 0

cos (θ) = - \frac{4}{9}

cos^{2} (x) + 2 cos (x) = sin^{2} (\frac{x}{2}) - 2

2 cos^{2} (x) - 4 cos (x) + 2 = 0

cos^{2} (x) - sin^{2} (x) = 1 - 2 sin (x)

What is the general solution for cos(3θ)=-1,0<= θ<2pi ?
The general solution for cos(3θ)=-1,0<= θ<2pi is θ= pi/3 ,θ=pi,θ=(5pi)/3