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

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

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

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-tan(2θ)+2sin(θ)=0,0<= θ<2pi

Solution

- tan (2 θ) + 2 sin (θ) = 0, 0 \leq θ < 2 π

Solution

θ = 0, θ = π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Degrees

θ = 0^{\circ}, θ = 18 0^{\circ}, θ = 12 0^{\circ}, θ = 24 0^{\circ}

Solution steps

- tan (2 θ) + 2 sin (θ) = 0, 0 \leq θ < 2 π

Express with sin, cos

- tan (2 θ) + 2 sin (θ)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - \frac{sin ( 2 θ )}{cos ( 2 θ )} + 2 sin (θ)

Simplify

- \frac{sin ( 2 θ )}{cos ( 2 θ )} + 2 sin (θ) : \frac{- sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

- \frac{sin ( 2 θ )}{cos ( 2 θ )} + 2 sin (θ)

Convert element to fraction:

2 sin (θ) = \frac{2 sin ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

= - \frac{sin ( 2 θ )}{cos ( 2 θ )} + \frac{2 sin ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

= \frac{- sin ( 2 θ ) + 2 sin ( θ ) cos ( 2 θ )}{cos ( 2 θ )}

\frac{- sin ( 2 θ ) + 2 cos ( 2 θ ) sin ( θ )}{cos ( 2 θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin (2 θ) + 2 cos (2 θ) sin (θ) = 0

Rewrite using trig identities

- sin (2 θ) + 2 cos (2 θ) sin (θ)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= - 2 sin (θ) cos (θ) + 2 cos (2 θ) sin (θ)

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ) = 0

Factor

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ) : 2 sin (θ) (cos (2 θ) - cos (θ))

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ)

Rewrite as

= 2 sin (θ) cos (2 θ) - 2 sin (θ) cos (θ)

Factor out common term

2 sin (θ)

= 2 sin (θ) (cos (2 θ) - cos (θ))

2 sin (θ) (cos (2 θ) - cos (θ)) = 0

Solving each part separately

sin (θ) = 0 or cos (2 θ) - cos (θ) = 0

sin (θ) = 0, 0 \leq θ < 2 π : θ = 0, θ = π

sin (θ) = 0, 0 \leq θ < 2 π

General solutions for

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = 0, θ = π

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

cos (2 θ) - cos (θ) = 0, 0 \leq θ < 2 π

Rewrite using trig identities

cos (2 θ) - cos (θ)

Use the Sum to Product identity:

cos (s) - cos (t) = - 2 sin (\frac{s + t}{2}) sin (\frac{s - t}{2})

= - 2 sin (\frac{2 θ + θ}{2}) sin (\frac{2 θ - θ}{2})

Simplify

- 2 sin (\frac{2 θ + θ}{2}) sin (\frac{2 θ - θ}{2}) : - 2 sin (\frac{3 θ}{2}) sin (\frac{θ}{2})

- 2 sin (\frac{2 θ + θ}{2}) sin (\frac{2 θ - θ}{2})

Add similar elements:

2 θ + θ = 3 θ

= - 2 sin (\frac{3 θ}{2}) sin (\frac{2 θ - θ}{2})

Add similar elements:

2 θ - θ = θ

= - 2 sin (\frac{3 θ}{2}) sin (\frac{θ}{2})

= - 2 sin (\frac{3 θ}{2}) sin (\frac{θ}{2})

- 2 sin (\frac{3 θ}{2}) sin (\frac{θ}{2}) = 0

Solving each part separately

sin (\frac{3 θ}{2}) = 0 or sin (\frac{θ}{2}) = 0

sin (\frac{3 θ}{2}) = 0, 0 \leq θ < 2 π : θ = 0, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

sin (\frac{3 θ}{2}) = 0, 0 \leq θ < 2 π

General solutions for

sin (\frac{3 θ}{2}) = 0

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}

\frac{3 θ}{2} = 0 + 2 πn, \frac{3 θ}{2} = π + 2 πn

\frac{3 θ}{2} = 0 + 2 πn, \frac{3 θ}{2} = π + 2 πn

Solve

\frac{3 θ}{2} = 0 + 2 πn : θ = \frac{4 πn}{3}

\frac{3 θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{3 θ}{2} = 2 πn

Multiply both sides by

2

\frac{3 θ}{2} = 2 πn

Multiply both sides by

2

\frac{2 \cdot 3 θ}{2} = 2 \cdot 2 πn

Simplify

3 θ = 4 πn

3 θ = 4 πn

Divide both sides by

3

3 θ = 4 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{4 πn}{3}

Simplify

θ = \frac{4 πn}{3}

θ = \frac{4 πn}{3}

Solve

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

\frac{3 θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{3 θ}{2} = π + 2 πn

Multiply both sides by

2

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

Simplify

3 θ = 2 π + 4 πn

3 θ = 2 π + 4 πn

Divide both sides by

3

3 θ = 2 π + 4 πn

Divide both sides by

3

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

Simplify

θ = \frac{2 π}{3} + \frac{4 πn}{3}

θ = \frac{2 π}{3} + \frac{4 πn}{3}

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

Solutions for the range

0 \leq θ < 2 π

θ = 0, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

sin (\frac{θ}{2}) = 0, 0 \leq θ < 2 π : θ = 0

sin (\frac{θ}{2}) = 0, 0 \leq θ < 2 π

General solutions for

sin (\frac{θ}{2}) = 0

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}

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

Solve

\frac{θ}{2} = 0 + 2 πn : θ = 4 πn

\frac{θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{θ}{2} = 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot 2 πn

Simplify

θ = 4 πn

θ = 4 πn

Solve

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

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{θ}{2} = π + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 π + 2 \cdot 2 πn

Simplify

θ = 2 π + 4 πn

θ = 2 π + 4 πn

θ = 4 πn, θ = 2 π + 4 πn

Solutions for the range

0 \leq θ < 2 π

θ = 0

Combine all the solutions

θ = 0, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Combine all the solutions

θ = 0, θ = π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Popular Examples

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cos (x) = \frac{2 tan ( x )}{1 + tan ^{2} ( x )}

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(1 + cos (4 x)) sin (2 x) = cos^{2} (2 x)

cos(2x)-cos(6x)-sin(4x)=0

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cot (θ) = \frac{15}{8}

sqrt(2)sin(2x)=sin(4x)

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Frequently Asked Questions (FAQ)

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