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

The general solution for tan(θ)=2sin(θ)tan(θ) is θ=pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(θ)=2sin(θ)tan(θ)

Solution

tan (θ) = 2 sin (θ) tan (θ)

Solution

θ = πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n, θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (θ) = 2 sin (θ) tan (θ)

Subtract

2 sin (θ) tan (θ)

from both sides

tan (θ) - 2 sin (θ) tan (θ) = 0

Factor

tan (θ) - 2 sin (θ) tan (θ) : - tan (θ) (2 sin (θ) - 1)

tan (θ) - 2 sin (θ) tan (θ)

Factor out common term

- tan (θ)

= - tan (θ) (- 1 + 2 sin (θ))

- tan (θ) (2 sin (θ) - 1) = 0

Solving each part separately

tan (θ) = 0 or 2 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

2 sin (θ) - 1 = 0 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

2 sin (θ) - 1 = 0

Move

1

to the right side

2 sin (θ) - 1 = 0

Add

1

to both sides

2 sin (θ) - 1 + 1 = 0 + 1

Simplify

2 sin (θ) = 1

2 sin (θ) = 1

Divide both sides by

2

2 sin (θ) = 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{1}{2}

Simplify

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

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}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Popular Examples

2sin^2(t)+sin(t)-1=0

2 sin^{2} (t) + sin (t) - 1 = 0

sin(θ)+sin(3θ)=0

sin (θ) + sin (3 θ) = 0

4sin^2(x)-1=0,0<= x<2pi

4 sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

cos^2(x)-4sin^2(x)=2

cos^{2} (x) - 4 sin^{2} (x) = 2

sin(x)=0.53

sin (x) = 0.53

Frequently Asked Questions (FAQ)

What is the general solution for tan(θ)=2sin(θ)tan(θ) ?
The general solution for tan(θ)=2sin(θ)tan(θ) is θ=pin,θ= pi/6+2pin,θ=(5pi)/6+2pin