English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4cos(β)sin(β)+2sin(β)-2cos(β)-1=0

Solution

4 cos (β) sin (β) + 2 sin (β) - 2 cos (β) - 1 = 0

Solution

β = \frac{2 π}{3} + 2 πn, β = \frac{4 π}{3} + 2 πn, β = \frac{π}{6} + 2 πn, β = \frac{5 π}{6} + 2 πn

Degrees

β = 12 0^{\circ} + 36 0^{\circ} n, β = 24 0^{\circ} + 36 0^{\circ} n, β = 3 0^{\circ} + 36 0^{\circ} n, β = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (β) sin (β) + 2 sin (β) - 2 cos (β) - 1 = 0

Factor

4 cos (β) sin (β) + 2 sin (β) - 2 cos (β) - 1 : (2 cos (β) + 1) (2 sin (β) - 1)

4 cos (β) sin (β) + 2 sin (β) - 2 cos (β) - 1

= (4 sin (β) cos (β) + 2 sin (β)) + (- 2 cos (β) - 1)

Factor out

- 1

from

- 2 cos (β) - 1 : - (2 cos (β) + 1)

- 2 cos (β) - 1

Factor out common term

- 1

= - (2 cos (β) + 1)

Factor out

2 sin (β)

from

4 sin (β) cos (β) + 2 sin (β) : 2 sin (β) (2 cos (β) + 1)

4 sin (β) cos (β) + 2 sin (β)

Rewrite

4

2 \cdot 2

= 2 \cdot 2 sin (β) cos (β) + 2 sin (β)

Factor out common term

2 sin (β)

= 2 sin (β) (2 cos (β) + 1)

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

Factor out common term

2 cos (β) + 1

= (2 cos (β) + 1) (2 sin (β) - 1)

(2 cos (β) + 1) (2 sin (β) - 1) = 0

Solving each part separately

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

2 cos (β) + 1 = 0 : β = \frac{2 π}{3} + 2 πn, β = \frac{4 π}{3} + 2 πn

2 cos (β) + 1 = 0

Move

1

to the right side

2 cos (β) + 1 = 0

Subtract

1

from both sides

2 cos (β) + 1 - 1 = 0 - 1

Simplify

2 cos (β) = - 1

2 cos (β) = - 1

Divide both sides by

2

2 cos (β) = - 1

Divide both sides by

2

\frac{2 cos ( β )}{2} = \frac{- 1}{2}

Simplify

cos (β) = - \frac{1}{2}

cos (β) = - \frac{1}{2}

General solutions for

cos (β) = - \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}

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

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

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

2 sin (β) - 1 = 0

Move

1

to the right side

2 sin (β) - 1 = 0

Add

1

to both sides

2 sin (β) - 1 + 1 = 0 + 1

Simplify

2 sin (β) = 1

2 sin (β) = 1

Divide both sides by

2

2 sin (β) = 1

Divide both sides by

2

\frac{2 sin ( β )}{2} = \frac{1}{2}

Simplify

sin (β) = \frac{1}{2}

sin (β) = \frac{1}{2}

General solutions for

sin (β) = \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}

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

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

Combine all the solutions

β = \frac{2 π}{3} + 2 πn, β = \frac{4 π}{3} + 2 πn, β = \frac{π}{6} + 2 πn, β = \frac{5 π}{6} + 2 πn

Popular Examples

5sin^2(x)=cos(x)

5 sin^{2} (x) = cos (x)

cos(x)=sqrt((1+cos(x))/2)

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

cos(x)=(sqrt(1+cos(x)))/2

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

sin(x/2)+(sqrt(3))/2 =0

sin (\frac{x}{2}) + \frac{3}{2} = 0

cos(x)=0.818

cos (x) = 0.818