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

The general solution for tan^2(θ)sin(θ)=tan^2(θ) is θ=pin

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tan^2(θ)sin(θ)=tan^2(θ)

tan^{2} (θ) sin (θ) = tan^{2} (θ)

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) sin (θ) = tan^{2} (θ)

Subtract

tan^{2} (θ)

from both sides

tan^{2} (θ) sin (θ) - tan^{2} (θ) = 0

Factor

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

tan^{2} (θ) sin (θ) - tan^{2} (θ)

Factor out common term

tan^{2} (θ)

= tan^{2} (θ) (sin (θ) - 1)

tan^{2} (θ) (sin (θ) - 1) = 0

Solving each part separately

tan^{2} (θ) = 0 or sin (θ) - 1 = 0

tan^{2} (θ) = 0 : θ = πn

tan^{2} (θ) = 0

Apply rule

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

tan (θ) = 0

General solutions for

tan (θ) = 0

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 = 0 + πn

θ = 0 + πn

Solve

θ = 0 + πn : θ = πn

θ = 0 + πn

0 + πn = πn

θ = πn

θ = πn

sin (θ) - 1 = 0 : θ = \frac{π}{2} + 2 πn

sin (θ) - 1 = 0

Move

1

to the right side

sin (θ) - 1 = 0

Add

1

to both sides

sin (θ) - 1 + 1 = 0 + 1

Simplify

sin (θ) = 1

sin (θ) = 1

General solutions for

sin (θ) = 1

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 = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = πn, θ = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

θ = πn

cos (x) = \frac{1}{cot ( x )}

(sin (x) - 1) \cdot (sin (x) - 3) = 0

2 - 3 cos (θ) = 0

2 sin (x) cos (x) = 0, 0 \leq x \leq 2 π

tan (θ) = - \frac{3}{4}, sec (θ)

What is the general solution for tan^2(θ)sin(θ)=tan^2(θ) ?
The general solution for tan^2(θ)sin(θ)=tan^2(θ) is θ=pin