What is the general solution for 2(tan(θ)+3)=5+tan(θ),0<= θ<2pi ?

Q: What is the general solution for 2(tan(θ)+3)=5+tan(θ),0<= θ<2pi ?

The general solution for 2(tan(θ)+3)=5+tan(θ),0<= θ<2pi is θ=(3pi)/4 ,θ=(7pi)/4

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Popular >

2(tan(θ)+3)=5+tan(θ),0<= θ<2pi

Solution

2 (tan (θ) + 3) = 5 + tan (θ), 0 \leq θ < 2 π

Solution

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 13 5^{\circ}, θ = 31 5^{\circ}

Solution steps

2 (tan (θ) + 3) = 5 + tan (θ), 0 \leq θ < 2 π

Solve by substitution

2 (tan (θ) + 3) = 5 + tan (θ)

Let:

tan (θ) = u

2 (u + 3) = 5 + u

2 (u + 3) = 5 + u : u = - 1

2 (u + 3) = 5 + u

Expand

2 (u + 3) : 2 u + 6

2 (u + 3)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = u, c = 3

= 2 u + 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 2 u + 6

2 u + 6 = 5 + u

Move

6

to the right side

2 u + 6 = 5 + u

Subtract

6

from both sides

2 u + 6 - 6 = 5 + u - 6

Simplify

2 u = u - 1

2 u = u - 1

Move

u

to the left side

2 u = u - 1

Subtract

u

from both sides

2 u - u = u - 1 - u

Simplify

u = - 1

u = - 1

Substitute back

u = tan (θ)

tan (θ) = - 1

tan (θ) = - 1

tan (θ) = - 1, 0 \leq θ < 2 π : θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

tan (θ) = - 1, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

Combine all the solutions

θ = \frac{3 π}{4}, θ = \frac{7 π}{4}

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Frequently Asked Questions (FAQ)

What is the general solution for 2(tan(θ)+3)=5+tan(θ),0<= θ<2pi ?
The general solution for 2(tan(θ)+3)=5+tan(θ),0<= θ<2pi is θ=(3pi)/4 ,θ=(7pi)/4