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

The general solution for 1+cos(a)=(2cos^2(a))/2 is a=2.23703…+2pin,a=-2.23703…+2pin

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1+cos(a)=(2cos^2(a))/2

Solution

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

Solution

a = 2.23703 \dots + 2 πn, a = - 2.23703 \dots + 2 πn

Degrees

a = 128.17270 \dots^{\circ} + 36 0^{\circ} n, a = - 128.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (a) = u

1 + u = \frac{2 u ^{2}}{2}

1 + u = \frac{2 u ^{2}}{2} : u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

1 + u = \frac{2 u ^{2}}{2}

Multiply both sides by

2

1 + u = \frac{2 u ^{2}}{2}

Multiply both sides by

2

1 \cdot 2 + u \cdot 2 = \frac{2 u ^{2}}{2} \cdot 2

Simplify

2 + 2 u = 2 u^{2}

2 + 2 u = 2 u^{2}

Switch sides

2 u^{2} = 2 + 2 u

Move

2 u

to the left side

2 u^{2} = 2 + 2 u

Subtract

2 u

from both sides

2 u^{2} - 2 u = 2 + 2 u - 2 u

Simplify

2 u^{2} - 2 u = 2

2 u^{2} - 2 u = 2

Move

2

to the left side

2 u^{2} - 2 u = 2

Subtract

2

from both sides

2 u^{2} - 2 u - 2 = 2 - 2

Simplify

2 u^{2} - 2 u - 2 = 0

2 u^{2} - 2 u - 2 = 0

Solve with the quadratic formula

2 u^{2} - 2 u - 2 = 0

Quadratic Equation Formula:

For

a = 2, b = - 2, c = - 2

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 (- 2) = 25

(- 2)^{2} - 4 \cdot 2 (- 2)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 2 \cdot 2

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

Add the numbers:

4 + 16 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

u_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 2}, u_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 2}

u = \frac{- ( - 2 ) + 2 5}{2 \cdot 2} : \frac{1 + 5}{2}

\frac{- ( - 2 ) + 2 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{2 + 2 5}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 + 2 5}{4}

Factor

2 + 25 : 2 (1 + 5)

2 + 25

Rewrite as

= 2 \cdot 1 + 25

Factor out common term

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{4}

Cancel the common factor:

2

= \frac{1 + 5}{2}

u = \frac{- ( - 2 ) - 2 5}{2 \cdot 2} : \frac{1 - 5}{2}

\frac{- ( - 2 ) - 2 5}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{2 - 2 5}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 - 2 5}{4}

Factor

2 - 25 : 2 (1 - 5)

2 - 25

Rewrite as

= 2 \cdot 1 - 25

Factor out common term

2

= 2 (1 - 5)

= \frac{2 ( 1 - 5 )}{4}

Cancel the common factor:

2

= \frac{1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

Substitute back

u = cos (a)

cos (a) = \frac{1 + 5}{2}, cos (a) = \frac{1 - 5}{2}

cos (a) = \frac{1 + 5}{2}, cos (a) = \frac{1 - 5}{2}

cos (a) = \frac{1 + 5}{2} :

No Solution

cos (a) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (a) = \frac{1 - 5}{2} : a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

cos (a) = \frac{1 - 5}{2}

Apply trig inverse properties

cos (a) = \frac{1 - 5}{2}

General solutions for

cos (a) = \frac{1 - 5}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

Combine all the solutions

a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

Show solutions in decimal form

a = 2.23703 \dots + 2 πn, a = - 2.23703 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1+cos(a)=(2cos^2(a))/2 ?
The general solution for 1+cos(a)=(2cos^2(a))/2 is a=2.23703…+2pin,a=-2.23703…+2pin