What is the general solution for 2cos(θ)-2=0 ?

The general solution for 2cos(θ)-2=0 is θ=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos(θ)-2=0

2 cos (θ) - 2 = 0

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (θ) - 2 = 0

Move

2

to the right side

2 cos (θ) - 2 = 0

Add

2

to both sides

2 cos (θ) - 2 + 2 = 0 + 2

Simplify

2 cos (θ) = 2

2 cos (θ) = 2

Divide both sides by

2

2 cos (θ) = 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{2}{2}

Simplify

cos (θ) = 1

cos (θ) = 1

General solutions for

cos (θ) = 1

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 = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

sin (x) - 2 = 0

x, sin^{4} (x) = \frac{cos ^{2} ( x )}{2}

tan^{2} (x) + 5 tan (x) - 1 = 0

csc (θ) + 1 = 0

2 sin^{2} (θ) - 3 sin (θ) + 1 = 0, 0 \leq θ \leq 2 π

What is the general solution for 2cos(θ)-2=0 ?
The general solution for 2cos(θ)-2=0 is θ=2pin