What is the general solution for tan(2θ)+tan(2θ)sin(2θ)=0 ?

The general solution for tan(2θ)+tan(2θ)sin(2θ)=0 is θ=(pin)/2

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tan(2θ)+tan(2θ)sin(2θ)=0

Solution

tan (2 θ) + tan (2 θ) sin (2 θ) = 0

Solution

θ = \frac{πn}{2}

Degrees

θ = 0^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 θ) + tan (2 θ) sin (2 θ) = 0

Factor

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

tan (2 θ) + tan (2 θ) sin (2 θ)

Factor out common term

tan (2 θ)

= tan (2 θ) (1 + sin (2 θ))

tan (2 θ) (sin (2 θ) + 1) = 0

Solving each part separately

tan (2 θ) = 0 or sin (2 θ) + 1 = 0

tan (2 θ) = 0 : θ = \frac{πn}{2}

tan (2 θ) = 0

General solutions for

tan (2 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}

2 θ = 0 + πn

2 θ = 0 + πn

Solve

2 θ = 0 + πn : θ = \frac{πn}{2}

2 θ = 0 + πn

0 + πn = πn

2 θ = πn

Divide both sides by

2

2 θ = πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{πn}{2}

Simplify

θ = \frac{πn}{2}

θ = \frac{πn}{2}

θ = \frac{πn}{2}

sin (2 θ) + 1 = 0 : θ = \frac{3 π}{4} + πn

sin (2 θ) + 1 = 0

Move

1

to the right side

sin (2 θ) + 1 = 0

Subtract

1

from both sides

sin (2 θ) + 1 - 1 = 0 - 1

Simplify

sin (2 θ) = - 1

sin (2 θ) = - 1

General solutions for

sin (2 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}

2 θ = \frac{3 π}{2} + 2 πn

2 θ = \frac{3 π}{2} + 2 πn

Solve

2 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{3 π}{4} + πn

θ = \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2θ)+tan(2θ)sin(2θ)=0 ?
The general solution for tan(2θ)+tan(2θ)sin(2θ)=0 is θ=(pin)/2