What is the general solution for tan(θ)sin(θ)-4tan(θ)=-3tan(θ) ?

The general solution for tan(θ)sin(θ)-4tan(θ)=-3tan(θ) is θ=pin

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tan(θ)sin(θ)-4tan(θ)=-3tan(θ)

tan (θ) sin (θ) - 4 tan (θ) = - 3 tan (θ)

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) sin (θ) - 4 tan (θ) = - 3 tan (θ)

Subtract

- 3 tan (θ)

from both sides

tan (θ) sin (θ) - tan (θ) = 0

Factor

tan (θ) sin (θ) - tan (θ) : tan (θ) (sin (θ) - 1)

tan (θ) sin (θ) - tan (θ)

Factor out common term

tan (θ)

= tan (θ) (sin (θ) - 1)

tan (θ) (sin (θ) - 1) = 0

Solving each part separately

tan (θ) = 0 or sin (θ) - 1 = 0

tan (θ) = 0 : θ = πn

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

sec^{2} (x) + tan (x) - 3 = 0

sin (x) = - 0.5

cos (x) = \frac{- 2}{2}

sin^{2} (θ) + sin (θ) = 0

cos^{2} (θ) - sin^{2} (θ) = sin (θ) + 1

What is the general solution for tan(θ)sin(θ)-4tan(θ)=-3tan(θ) ?
The general solution for tan(θ)sin(θ)-4tan(θ)=-3tan(θ) is θ=pin