What is the general solution for 1+cos(a)=(3+4cos(a))/2 ?

The general solution for 1+cos(a)=(3+4cos(a))/2 is a=(2pi)/3+2pin,a=(4pi)/3+2pin

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Popular >

1+cos(a)=(3+4cos(a))/2

Solution

1 + cos (a) = \frac{3 + 4 cos ( a )}{2}

Solution

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

Degrees

a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + cos (a) = \frac{3 + 4 cos ( a )}{2}

Solve by substitution

1 + cos (a) = \frac{3 + 4 cos ( a )}{2}

Let:

cos (a) = u

1 + u = \frac{3 + 4 u}{2}

1 + u = \frac{3 + 4 u}{2} : u = - \frac{1}{2}

1 + u = \frac{3 + 4 u}{2}

Multiply both sides by

2

1 + u = \frac{3 + 4 u}{2}

Multiply both sides by

2

1 \cdot 2 + u \cdot 2 = \frac{3 + 4 u}{2} \cdot 2

Simplify

1 \cdot 2 + u \cdot 2 = \frac{3 + 4 u}{2} \cdot 2

Simplify

1 \cdot 2 : 2

1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 2

Simplify

u \cdot 2 : 2 u

u \cdot 2

Apply the commutative law:

u \cdot 2 = 2 u

2 u

Simplify

\frac{3 + 4 u}{2} \cdot 2 : 3 + 4 u

\frac{3 + 4 u}{2} \cdot 2

Multiply fractions:

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

= \frac{( 3 + 4 u ) \cdot 2}{2}

Cancel the common factor:

2

= 3 + 4 u

2 + 2 u = 3 + 4 u

2 + 2 u = 3 + 4 u

2 + 2 u = 3 + 4 u

Move

2

to the right side

2 + 2 u = 3 + 4 u

Subtract

2

from both sides

2 + 2 u - 2 = 3 + 4 u - 2

Simplify

2 u = 4 u + 1

2 u = 4 u + 1

Move

4 u

to the left side

2 u = 4 u + 1

Subtract

4 u

from both sides

2 u - 4 u = 4 u + 1 - 4 u

Simplify

- 2 u = 1

- 2 u = 1

Divide both sides by

- 2

- 2 u = 1

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{1}{- 2}

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Substitute back

u = cos (a)

cos (a) = - \frac{1}{2}

cos (a) = - \frac{1}{2}

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

cos (a) = - \frac{1}{2}

General solutions for

cos (a) = - \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}

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

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

Combine all the solutions

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

Popular Examples

sqrt(3)cos(x)-2cos^2(x)=0,0<x<360

3 cos (x) - 2 cos^{2} (x) = 0, 0^{\circ} < x < 36 0^{\circ}

sin^2(x)cos(x)-sin(x)cos^3(x)=0

sin^{2} (x) cos (x) - sin (x) cos^{3} (x) = 0

5sin(θ)-sqrt(3)=3sin(θ)

5 sin (θ) - 3 = 3 sin (θ)

sin^3(x)-4sin(x)=0

sin^{3} (x) - 4 sin (x) = 0

cos(a)=(-82)/(sqrt(140)*\sqrt{138)}

cos (a) = \frac{- 82}{140 \cdot 138}

Frequently Asked Questions (FAQ)

What is the general solution for 1+cos(a)=(3+4cos(a))/2 ?
The general solution for 1+cos(a)=(3+4cos(a))/2 is a=(2pi)/3+2pin,a=(4pi)/3+2pin