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

The general solution for sin(2θ)=(sqrt(3))/2 ,0<= θ<2pi is θ= pi/6 ,θ= pi/3 ,θ=(7pi)/6 ,θ=(4pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2θ)=(sqrt(3))/2 ,0<= θ<2pi

sin (2 θ) = \frac{3}{2}, 0 \leq θ < 2 π

θ = \frac{π}{6}, θ = \frac{π}{3}, θ = \frac{7 π}{6}, θ = \frac{4 π}{3}

Solution steps

sin (2 θ) = \frac{3}{2}, 0 \leq θ < 2 π

General solutions for

sin (2 θ) = \frac{3}{2}

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

Solve

2 θ = \frac{π}{3} + 2 πn : θ = \frac{π}{6} + πn

Solve

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

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{6}, θ = \frac{π}{3}, θ = \frac{7 π}{6}, θ = \frac{4 π}{3}

- sin (4 x) = 0

csc (x) = sec (4 1^{\circ})

sin (x) - sin (3 x) = 0

2 sin^{2} (θ) - 5 cos (θ) + 1 = 0

sec^{2} (x) cos (x) = cos (x), 0^{\circ} < x < 36 0^{\circ}

What is the general solution for sin(2θ)=(sqrt(3))/2 ,0<= θ<2pi ?
The general solution for sin(2θ)=(sqrt(3))/2 ,0<= θ<2pi is θ= pi/6 ,θ= pi/3 ,θ=(7pi)/6 ,θ=(4pi)/3