What is the general solution for 2sin(piθ)=1 ?

The general solution for 2sin(piθ)=1 is θ= 1/6+2n,θ= 5/6+2n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(piθ)=1

2 sin (π θ) = 1

θ = \frac{1}{6} + 2 n, θ = \frac{5}{6} + 2 n

Solution steps

2 sin (π θ) = 1

Divide both sides by

2

sin (π θ) = \frac{1}{2}

General solutions for

sin (π θ) = \frac{1}{2}

π θ = \frac{π}{6} + 2 πn, π θ = \frac{5 π}{6} + 2 πn

Solve

π θ = \frac{π}{6} + 2 πn : θ = \frac{1}{6} + 2 n

Solve

π θ = \frac{5 π}{6} + 2 πn : θ = \frac{5}{6} + 2 n

θ = \frac{1}{6} + 2 n, θ = \frac{5}{6} + 2 n

23 sin (\frac{x}{2}) = 3

2 cos (θ) = - 1

sin (θ) = \frac{2}{5}

cos (2 x) - cos (x) = 0, 0 \leq x \leq 2 π

cos (θ) = 2

What is the general solution for 2sin(piθ)=1 ?
The general solution for 2sin(piθ)=1 is θ= 1/6+2n,θ= 5/6+2n