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

The general solution for 2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin

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Popular >

2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1

Solution

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

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Factor

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

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

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

Factor out

- 1

from

- 2 sin (θ) - 1 : - (2 sin (θ) + 1)

- 2 sin (θ) - 1

Factor out common term

- 1

= - (2 sin (θ) + 1)

Factor out

cos (θ)

from

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

2 sin (θ) cos (θ) + cos (θ)

Factor out common term

cos (θ)

= cos (θ) (2 sin (θ) + 1)

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

Factor out common term

2 sin (θ) + 1

= (2 sin (θ) + 1) (cos (θ) - 1)

(2 sin (θ) + 1) (cos (θ) - 1) = 0

Solving each part separately

2 sin (θ) + 1 = 0 or cos (θ) - 1 = 0

2 sin (θ) + 1 = 0 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

2 sin (θ) + 1 = 0

Move

1

to the right side

2 sin (θ) + 1 = 0

Subtract

1

from 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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cos (θ) - 1 = 0 : θ = 2 πn

cos (θ) - 1 = 0

Move

1

to the right side

cos (θ) - 1 = 0

Add

1

to both sides

cos (θ) - 1 + 1 = 0 + 1

Simplify

cos (θ) = 1

cos (θ) = 1

General solutions for

cos (θ) = 1

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 = 0 + 2 πn

θ = 0 + 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

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn

Popular Examples

sin(2θ)-cos(θ)-2sin(θ)+1=0

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

2cos(x)+1=0-2,pi<= x<= 2pi

2 cos (x) + 1 = 0 - 2, π \leq x \leq 2 π

cos^3(x)=-cos(x)

cos^{3} (x) = - cos (x)

csc(2x)=2sin(2x),0<= x<= 2pi

csc (2 x) = 2 sin (2 x), 0 \leq x \leq 2 π

cos(x)=-12/13

cos (x) = - \frac{12}{13}

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1 ?
The general solution for 2cos(θ)sin(θ)+cos(θ)-2sin(θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin