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

The general solution for 2sin^2(2θ)=1-sin(2θ) is θ= pi/(12)+pin,θ=(5pi}{12}+pin,θ=\frac{3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(2θ)=1-sin(2θ)

2 sin^{2} (2 θ) = 1 - sin (2 θ)

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn, θ = \frac{3 π}{4} + πn

Solution steps

2 sin^{2} (2 θ) = 1 - sin (2 θ)

Solve by substitution

sin (2 θ) = \frac{1}{2}, sin (2 θ) = - 1

sin (2 θ) = \frac{1}{2} : θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

sin (2 θ) = - 1 : θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn, θ = \frac{3 π}{4} + πn

cos (θ) = \frac{24}{25}

sec^{2} (x) = 1 - tan^{2} (x)

8 sin^{2} (θ) - 17 sin (θ) + 9 = 0

sin^{2} (x) = 2 cos^{2} (x)

9 sin (2 θ) - 2 sin (θ) = 0

What is the general solution for 2sin^2(2θ)=1-sin(2θ) ?
The general solution for 2sin^2(2θ)=1-sin(2θ) is θ= pi/(12)+pin,θ=(5pi}{12}+pin,θ=\frac{3pi)/4+pin