What is the general solution for 2sin(θ)=sqrt(2),0<= θ<2pi ?

The general solution for 2sin(θ)=sqrt(2),0<= θ<2pi is θ= pi/4 ,θ=(3pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(θ)=sqrt(2),0<= θ<2pi

2 sin (θ) = 2, 0 \leq θ < 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

Solution steps

2 sin (θ) = 2, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

2 cos^{2} (t) - 3 cos (t) = - 1

cos (t) - cos (2 t) = - 0.75

2 cos (A) = - 3

4 sin (x) cos (x) = cos (x)

cos (x) = 2 sin (x) cos (x)

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