What is the general solution for 1-sqrt(2)sin(θ)=0 ?

The general solution for 1-sqrt(2)sin(θ)=0 is θ= pi/4+2pin,θ=(3pi)/4+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1-sqrt(2)sin(θ)=0

1 - 2 sin (θ) = 0

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

Solution steps

1 - 2 sin (θ) = 0

Move

1

to the right side

- 2 sin (θ) = - 1

Divide both sides by

- 2

sin (θ) = \frac{2}{2}

General solutions for

sin (θ) = \frac{2}{2}

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

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

2 cos^{2} (θ) + 4 sin (θ) - 13 = 9 sin (θ) - 9

cot (θ) = - \frac{5}{4}

tan^{2} (θ) + 5 tan (θ) = 0

sin (4 x) - sin (6 x) = 0

What is the general solution for 1-sqrt(2)sin(θ)=0 ?
The general solution for 1-sqrt(2)sin(θ)=0 is θ= pi/4+2pin,θ=(3pi)/4+2pin