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

The general solution for 2sin(2θ)=-1 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(2θ)=-1

2 sin (2 θ) = - 1

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Solution steps

2 sin (2 θ) = - 1

Divide both sides by

2

sin (2 θ) = - \frac{1}{2}

General solutions for

sin (2 θ) = - \frac{1}{2}

2 θ = \frac{7 π}{6} + 2 πn, 2 θ = \frac{11 π}{6} + 2 πn

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

Solve

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{12} + πn

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

sin (4 θ) = - \frac{2 3}{3}

cos (2 x) + 5 sin (x) + 2 = 0

2 sin^{2} (θ) - 3 cos (θ) = 3

cot (θ) = \frac{1}{3}

sec (x) csc (x) = 2 sec (x)

What is the general solution for 2sin(2θ)=-1 ?
The general solution for 2sin(2θ)=-1 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin