What is the general solution for 2sin(θ)tan(θ)-tan(θ)=1-2sin(θ) ?

The general solution for 2sin(θ)tan(θ)-tan(θ)=1-2sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/4+pin

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Popular >

2sin(θ)tan(θ)-tan(θ)=1-2sin(θ)

Solution

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

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1 - 2 sin (θ)

from both sides

2 sin (θ) tan (θ) - tan (θ) - 1 + 2 sin (θ) = 0

Factor

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

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

= (2 sin (θ) tan (θ) - tan (θ)) + (2 sin (θ) - 1)

Factor out

tan (θ)

from

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

2 sin (θ) tan (θ) - tan (θ)

Factor out common term

tan (θ)

= tan (θ) (2 sin (θ) - 1)

= (2 sin (θ) - 1) + tan (θ) (2 sin (θ) - 1)

Factor out common term

2 sin (θ) - 1

= (2 sin (θ) - 1) (tan (θ) + 1)

(2 sin (θ) - 1) (tan (θ) + 1) = 0

Solving each part separately

2 sin (θ) - 1 = 0 or tan (θ) + 1 = 0

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

tan (θ) + 1 = 0 : θ = \frac{3 π}{4} + πn

tan (θ) + 1 = 0

Move

1

to the right side

tan (θ) + 1 = 0

Subtract

1

from 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{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(θ)tan(θ)-tan(θ)=1-2sin(θ) ?
The general solution for 2sin(θ)tan(θ)-tan(θ)=1-2sin(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/4+pin