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

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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}

Degrees

θ = 0^{\circ}, θ = 18 0^{\circ}, θ = 9 0^{\circ}, θ = 27 0^{\circ}

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

sin^{2} (θ) = 0, 0 \leq θ < 2 π

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Solutions for the range

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}

cos^{2} (θ) = 0, 0 \leq θ < 2 π

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

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