What is the general solution for 2sin(x)cos(x)+2sin(x)-cos(x)-1=0 ?

The general solution for 2sin(x)cos(x)+2sin(x)-cos(x)-1=0 is x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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Popular >

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

Solution

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

Solution

x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Factor

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

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

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

Factor out

- 1

from

- cos (x) - 1 : - (cos (x) + 1)

- cos (x) - 1

Factor out common term

- 1

= - (cos (x) + 1)

Factor out

2 sin (x)

from

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

2 sin (x) cos (x) + 2 sin (x)

Factor out common term

2 sin (x)

= 2 sin (x) (cos (x) + 1)

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

Factor out common term

cos (x) + 1

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

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

Solving each part separately

cos (x) + 1 = 0 or 2 sin (x) - 1 = 0

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

cos (x) + 1 - 1 = 0 - 1

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 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}

x = π + 2 πn

x = π + 2 πn

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

2 sin (x) - 1 = 0

Move

1

to the right side

2 sin (x) - 1 = 0

Add

1

to both sides

2 sin (x) - 1 + 1 = 0 + 1

Simplify

2 sin (x) = 1

2 sin (x) = 1

Divide both sides by

2

2 sin (x) = 1

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{1}{2}

Simplify

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \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}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

sin(θ)tan^2(θ)-sin(θ)=0

sin (θ) tan^{2} (θ) - sin (θ) = 0

sin(x)= 4/5 ,sin(180-x)

sin (x) = \frac{4}{5}, sin (18 0^{\circ} - x)

-11cos(x)(22)-11sin(x)(8)=-250

- 11 cos (x) (22) - 11 sin (x) (8) = - 250

3cos(x)+1=2sin(x)

3 cos (x) + 1 = 2 sin (x)

pi/6 =cos(2x)

\frac{π}{6} = cos (2 x)

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)cos(x)+2sin(x)-cos(x)-1=0 ?
The general solution for 2sin(x)cos(x)+2sin(x)-cos(x)-1=0 is x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin