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

The general solution for sin(θ)=(sqrt(2))/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 >

sin(θ)=(sqrt(2))/2 ,0<= θ<= 2pi

sin (θ) = \frac{2}{2}, 0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

Degrees

θ = 4 5^{\circ}, θ = 13 5^{\circ}

Solution steps

sin (θ) = \frac{2}{2}, 0 \leq θ \leq 2 π

General solutions for

sin (θ) = \frac{2}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

x, cos (x) = \frac{2}{3}

3 sin (x) + 2 = 1.5

3 sec^{2} (x) - 2 tan (x) - 4 = 0

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

sin (x) = \frac{1}{3.5}

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