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

The general solution for 5cos(θ)=2+cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5cos(θ)=2+cos(θ)

5 cos (θ) = 2 + cos (θ)

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

5 cos (θ) = 2 + cos (θ)

Solve by substitution

5 cos (θ) = 2 + cos (θ)

Let:

cos (θ) = u

5 u = 2 + u

5 u = 2 + u : u = \frac{1}{2}

5 u = 2 + u

Move

u

to the left side

5 u = 2 + u

Subtract

u

from both sides

5 u - u = 2 + u - u

Simplify

4 u = 2

4 u = 2

Divide both sides by

4

4 u = 2

Divide both sides by

4

\frac{4 u}{4} = \frac{2}{4}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (x + π) = 1

x, sin (4 x) = - 2 (sin (2 x))

(1 + cos (2 x)) sin (x) = 2 cos (x) sin (x)

sin (b) = \frac{2}{3}

8 tan (x) - 3 = 4

What is the general solution for 5cos(θ)=2+cos(θ) ?
The general solution for 5cos(θ)=2+cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin