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

The general solution for sin(θ)=2sin(θ)cos(θ) is θ=2pin,θ=pi+2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin

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Popular >

sin(θ)=2sin(θ)cos(θ)

Solution

sin (θ) = 2 sin (θ) cos (θ)

Solution

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

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) = 2 sin (θ) cos (θ)

Subtract

2 sin (θ) cos (θ)

from both sides

sin (θ) - 2 sin (θ) cos (θ) = 0

Factor

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

sin (θ) - 2 sin (θ) cos (θ)

Factor out common term

- sin (θ)

= - sin (θ) (- 1 + 2 cos (θ))

- sin (θ) (2 cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (θ) - 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

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

2 cos (θ) - 1 = 0 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

2 cos (θ) - 1 = 0

Move

1

to the right side

2 cos (θ) - 1 = 0

Add

1

to both sides

2 cos (θ) - 1 + 1 = 0 + 1

Simplify

2 cos (θ) = 1

2 cos (θ) = 1

Divide both sides by

2

2 cos (θ) = 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

Simplify

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Combine all the solutions

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

Popular Examples

solvefor x,arccos(e^{-x})= pi/3 solve for

x, arccos (e^{- x}) = \frac{π}{3}

2cos^2(4x+1)=0

2 cos^{2} (4 x + 1) = 0

solvefor x,2sin(x)=-1 solve for

x, 2 sin (x) = - 1

solvefor t,f=acos((tpi)/6-(11pi)/(12))solve for

t, f = a cos (\frac{t π}{6} - \frac{11 π}{12})

-9cos^2(θ)+9=16sin(θ)-7,0<= θ<360

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)=2sin(θ)cos(θ) ?
The general solution for sin(θ)=2sin(θ)cos(θ) is θ=2pin,θ=pi+2pin,θ= pi/3+2pin,θ=(5pi)/3+2pin