What is the general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 ?

The general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 is x=2.53724…+2pin,x=-2.53724…+2pin

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2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2

Solution

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

Solution

x = 2.53724 \dots + 2 πn, x = - 2.53724 \dots + 2 πn

Degrees

x = 145.37370 \dots^{\circ} + 36 0^{\circ} n, x = - 145.37370 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 + cos (x) = 3 \cdot \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

Subtract

3 \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

from both sides

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

Simplify

2 + cos (x) - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2} : \frac{4 + 2 cos ( x ) - 3 cos ^{2} ( x ) - sin ^{2} ( x )}{2}

2 + cos (x) - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2}

Convert element to fraction:

2 = \frac{2 \cdot 2}{2}, cos (x) = \frac{cos ( x ) 2}{2}

= \frac{2 \cdot 2}{2} + \frac{cos ( x ) \cdot 2}{2} - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 \cdot 2 + cos ( x ) \cdot 2 - ( 3 cos ^{2} ( x ) + sin ^{2} ( x ) )}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 + 2 cos ( x ) - ( 3 cos ^{2} ( x ) + sin ^{2} ( x ) )}{2}

Expand

4 + cos (x) \cdot 2 - (3 cos^{2} (x) + sin^{2} (x)) : 4 + cos (x) \cdot 2 - 3 cos^{2} (x) - sin^{2} (x)

4 + cos (x) \cdot 2 - (3 cos^{2} (x) + sin^{2} (x))

= 4 + 2 cos (x) - (3 cos^{2} (x) + sin^{2} (x))

- (3 cos^{2} (x) + sin^{2} (x)) : - 3 cos^{2} (x) - sin^{2} (x)

- (3 cos^{2} (x) + sin^{2} (x))

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

= 4 + cos (x) \cdot 2 - 3 cos^{2} (x) - sin^{2} (x)

= \frac{4 + 2 cos ( x ) - 3 cos ^{2} ( x ) - sin ^{2} ( x )}{2}

\frac{4 + 2 cos ( x ) - 3 cos ^{2} ( x ) - sin ^{2} ( x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 + 2 cos (x) - 3 cos^{2} (x) - sin^{2} (x) = 0

Rewrite using trig identities

4 - sin^{2} (x) + 2 cos (x) - 3 cos^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 4 - (1 - cos^{2} (x)) + 2 cos (x) - 3 cos^{2} (x)

Simplify

4 - (1 - cos^{2} (x)) + 2 cos (x) - 3 cos^{2} (x) : 2 cos (x) - 2 cos^{2} (x) + 3

4 - (1 - cos^{2} (x)) + 2 cos (x) - 3 cos^{2} (x)

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

- (1 - cos^{2} (x))

Distribute parentheses

= - (1) - (- cos^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (x)

= 4 - 1 + cos^{2} (x) + 2 cos (x) - 3 cos^{2} (x)

Simplify

4 - 1 + cos^{2} (x) + 2 cos (x) - 3 cos^{2} (x) : 2 cos (x) - 2 cos^{2} (x) + 3

4 - 1 + cos^{2} (x) + 2 cos (x) - 3 cos^{2} (x)

Group like terms

= cos^{2} (x) + 2 cos (x) - 3 cos^{2} (x) + 4 - 1

Add similar elements:

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

= - 2 cos^{2} (x) + 2 cos (x) + 4 - 1

Add/Subtract the numbers:

4 - 1 = 3

= 2 cos (x) - 2 cos^{2} (x) + 3

= 2 cos (x) - 2 cos^{2} (x) + 3

= 2 cos (x) - 2 cos^{2} (x) + 3

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

Solve by substitution

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

Let:

cos (x) = u

3 + 2 u - 2 u^{2} = 0

3 + 2 u - 2 u^{2} = 0 : u = - \frac{- 1 + 7}{2}, u = \frac{1 + 7}{2}

3 + 2 u - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + 2 u + 3 = 0

Solve with the quadratic formula

- 2 u^{2} + 2 u + 3 = 0

Quadratic Equation Formula:

For

a = - 2, b = 2, c = 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

2^{2} - 4 (- 2) \cdot 3 = 27

2^{2} - 4 (- 2) \cdot 3

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 2^{2} + 24

2^{2} = 4

= 4 + 24

Add the numbers:

4 + 24 = 28

= 28

Prime factorization of

28 : 2^{2} \cdot 7

28

28

divides by

228 = 14 \cdot 2

= 2 \cdot 14

14

divides by

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

Apply radical rule:

= 7 2^{2}

Apply radical rule:

2^{2} = 2

= 27

u_{1, 2} = \frac{- 2 \pm 2 7}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 2 + 2 7}{2 ( - 2 )}, u_{2} = \frac{- 2 - 2 7}{2 ( - 2 )}

u = \frac{- 2 + 2 7}{2 ( - 2 )} : - \frac{- 1 + 7}{2}

\frac{- 2 + 2 7}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 2 7}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 + 2 7}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 2 + 2 7}{4}

Cancel

\frac{- 2 + 2 7}{4} : \frac{7 - 1}{2}

\frac{- 2 + 2 7}{4}

Factor

- 2 + 27 : 2 (- 1 + 7)

- 2 + 27

Rewrite as

= - 2 \cdot 1 + 27

Factor out common term

2

= 2 (- 1 + 7)

= \frac{2 ( - 1 + 7 )}{4}

Cancel the common factor:

2

= \frac{- 1 + 7}{2}

= - \frac{7 - 1}{2}

= - \frac{- 1 + 7}{2}

u = \frac{- 2 - 2 7}{2 ( - 2 )} : \frac{1 + 7}{2}

\frac{- 2 - 2 7}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 2 7}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2 - 2 7}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

- 2 - 27 = - (2 + 27)

= \frac{2 + 2 7}{4}

Factor

2 + 27 : 2 (1 + 7)

2 + 27

Rewrite as

= 2 \cdot 1 + 27

Factor out common term

2

= 2 (1 + 7)

= \frac{2 ( 1 + 7 )}{4}

Cancel the common factor:

2

= \frac{1 + 7}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 7}{2}, u = \frac{1 + 7}{2}

Substitute back

u = cos (x)

cos (x) = - \frac{- 1 + 7}{2}, cos (x) = \frac{1 + 7}{2}

cos (x) = - \frac{- 1 + 7}{2}, cos (x) = \frac{1 + 7}{2}

cos (x) = - \frac{- 1 + 7}{2} : x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

cos (x) = - \frac{- 1 + 7}{2}

Apply trig inverse properties

cos (x) = - \frac{- 1 + 7}{2}

General solutions for

cos (x) = - \frac{- 1 + 7}{2}

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

x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

cos (x) = \frac{1 + 7}{2} :

No Solution

cos (x) = \frac{1 + 7}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

Show solutions in decimal form

x = 2.53724 \dots + 2 πn, x = - 2.53724 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 ?
The general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 is x=2.53724…+2pin,x=-2.53724…+2pin