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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(θ)-sqrt(3)=0,0<= θ<= 2pi

2 sin (θ) - 3 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{2 π}{3}

Solution steps

2 sin (θ) - 3 = 0, 0 \leq θ \leq 2 π

Move

3

to the right side

2 sin (θ) = 3

Divide both sides by

2

sin (θ) = \frac{3}{2}

General solutions for

sin (θ) = \frac{3}{2}

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{2 π}{3}

1 sin (x) tan (x) + 3 sin (x) = 0

6 (sin (x))^{2} - 13 cos (x) = 11

0 = cos (x) - sin (2 x)

tan^{2} (x) + 2 + 2 tan (x) - 7 = 0

(2 cos (x) + 1) = 0

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