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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^2(θ)cos^2(θ)=0,0<= θ<2pi

sin^{2} (θ) cos^{2} (θ) = 0, 0 \leq θ < 2 π

θ = 0, θ = π, θ = \frac{π}{2}, θ = \frac{3 π}{2}

Solution steps

sin^{2} (θ) cos^{2} (θ) = 0, 0 \leq θ < 2 π

Solving each part separately

sin^{2} (θ) = 0 or cos^{2} (θ) = 0

sin^{2} (θ) = 0, 0 \leq θ < 2 π : θ = 0, θ = π

cos^{2} (θ) = 0, 0 \leq θ < 2 π : θ = \frac{π}{2}, θ = \frac{3 π}{2}

Combine all the solutions

θ = 0, θ = π, θ = \frac{π}{2}, θ = \frac{3 π}{2}

tan (θ) + 2 = 0

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

sin (2 x + \frac{π}{3}) = 0

300 = \frac{1}{32} 10 0^{2} sin (2 x)

sin^{2} (x) = \frac{45}{52}

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