What is the general solution for 4sin(θ)+1=2sin(θ) ?

The general solution for 4sin(θ)+1=2sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4sin(θ)+1=2sin(θ)

4 sin (θ) + 1 = 2 sin (θ)

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

4 sin (θ) + 1 = 2 sin (θ)

Solve by substitution

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cot (θ) = \frac{1}{2}

sin (θ) = 0.7

sin (θ) = 0.2

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

2 sin (x) = - 1

What is the general solution for 4sin(θ)+1=2sin(θ) ?
The general solution for 4sin(θ)+1=2sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin