What is the general solution for 2sin^2(θ)tan(θ)-2sin^2(θ)=0 ?

The general solution for 2sin^2(θ)tan(θ)-2sin^2(θ)=0 is θ=2pin,θ=pi+2pin,θ= pi/4+pin

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2sin^2(θ)tan(θ)-2sin^2(θ)=0

2 sin^{2} (θ) tan (θ) - 2 sin^{2} (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{4} + πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

2 sin^{2} (θ) tan (θ) - 2 sin^{2} (θ) = 0

Factor

2 sin^{2} (θ) tan (θ) - 2 sin^{2} (θ) : 2 sin^{2} (θ) (tan (θ) - 1)

2 sin^{2} (θ) tan (θ) - 2 sin^{2} (θ)

Factor out common term

2 sin^{2} (θ)

= 2 sin^{2} (θ) (tan (θ) - 1)

2 sin^{2} (θ) (tan (θ) - 1) = 0

Solving each part separately

sin^{2} (θ) = 0 or tan (θ) - 1 = 0

sin^{2} (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin^{2} (θ) = 0

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

tan (θ) - 1 = 0 : θ = \frac{π}{4} + πn

tan (θ) - 1 = 0

Move

1

to the right side

tan (θ) - 1 = 0

Add

1

to both sides

tan (θ) - 1 + 1 = 0 + 1

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{4} + πn

15 cos^{2} (θ) + 2 sin^{2} (θ) = 7

x, csc (x) + cot (x) = \frac{3}{3}

5 cos (θ) + 1 = 8 cos (θ)

tan (x) = - \frac{12}{35}

cos (x) sin (x) = cos (x)

What is the general solution for 2sin^2(θ)tan(θ)-2sin^2(θ)=0 ?
The general solution for 2sin^2(θ)tan(θ)-2sin^2(θ)=0 is θ=2pin,θ=pi+2pin,θ= pi/4+pin