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

The general solution for 3cos^2(x)-2sqrt(3)cos(x)=sin^2(x) is x=1.80125…+2pin,x=-1.80125…+2pin

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

Solution

3 cos^{2} (x) - 23 cos (x) = sin^{2} (x)

Solution

x = 1.80125 \dots + 2 πn, x = - 1.80125 \dots + 2 πn

Degrees

x = 103.20437 \dots^{\circ} + 36 0^{\circ} n, x = - 103.20437 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (x) - 23 cos (x) = sin^{2} (x)

Subtract

sin^{2} (x)

from both sides

3 cos^{2} (x) - 23 cos (x) - sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- (1 - cos^{2} (x)) + 3 cos^{2} (x) - 2 cos (x) 3 : - 1 + 4 cos^{2} (x) - 23 cos (x)

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

= - (1 - cos^{2} (x)) + 3 cos^{2} (x) - 23 cos (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)

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

Add similar elements:

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

= - 1 + 4 cos^{2} (x) - 23 cos (x)

= - 1 + 4 cos^{2} (x) - 23 cos (x)

- 1 + 4 cos^{2} (x) - 2 cos (x) 3 = 0

Solve by substitution

- 1 + 4 cos^{2} (x) - 2 cos (x) 3 = 0

Let:

cos (x) = u

- 1 + 4 u^{2} - 2 u 3 = 0

- 1 + 4 u^{2} - 2 u 3 = 0 : u = \frac{3 + 7}{4}, u = \frac{3 - 7}{4}

- 1 + 4 u^{2} - 2 u 3 = 0

Write in the standard form

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

4 u^{2} - 23 u - 1 = 0

Solve with the quadratic formula

4 u^{2} - 23 u - 1 = 0

Quadratic Equation Formula:

For

a = 4, b = - 23, c = - 1

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

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

(- 23)^{2} - 4 \cdot 4 (- 1) = 27

(- 23)^{2} - 4 \cdot 4 (- 1)

Apply rule

- (- a) = a

= (- 23)^{2} + 4 \cdot 4 \cdot 1

(- 23)^{2} = 2^{2} \cdot 3

(- 23)^{2}

Apply exponent rule:

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

n

is even

(- 23)^{2} = (23)^{2}

= (23)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (3)^{2}

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 2^{2} \cdot 3

4 \cdot 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 16

= 2^{2} \cdot 3 + 16

2^{2} \cdot 3 = 12

2^{2} \cdot 3

2^{2} = 4

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 12 + 16

Add the numbers:

12 + 16 = 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 3 ) \pm 2 7}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 2 3 ) + 2 7}{2 \cdot 4}, u_{2} = \frac{- ( - 2 3 ) - 2 7}{2 \cdot 4}

u = \frac{- ( - 2 3 ) + 2 7}{2 \cdot 4} : \frac{3 + 7}{4}

\frac{- ( - 2 3 ) + 2 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 3 + 2 7}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 3 + 2 7}{8}

Factor out common term

2

= \frac{2 ( 3 + 7 )}{8}

Cancel the common factor:

2

= \frac{3 + 7}{4}

u = \frac{- ( - 2 3 ) - 2 7}{2 \cdot 4} : \frac{3 - 7}{4}

\frac{- ( - 2 3 ) - 2 7}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 3 - 2 7}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 3 - 2 7}{8}

Factor out common term

2

= \frac{2 ( 3 - 7 )}{8}

Cancel the common factor:

2

= \frac{3 - 7}{4}

The solutions to the quadratic equation are:

u = \frac{3 + 7}{4}, u = \frac{3 - 7}{4}

Substitute back

u = cos (x)

cos (x) = \frac{3 + 7}{4}, cos (x) = \frac{3 - 7}{4}

cos (x) = \frac{3 + 7}{4}, cos (x) = \frac{3 - 7}{4}

cos (x) = \frac{3 + 7}{4} :

No Solution

cos (x) = \frac{3 + 7}{4}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{3 - 7}{4} : x = arccos (\frac{3 - 7}{4}) + 2 πn, x = - arccos (\frac{3 - 7}{4}) + 2 πn

cos (x) = \frac{3 - 7}{4}

Apply trig inverse properties

cos (x) = \frac{3 - 7}{4}

General solutions for

cos (x) = \frac{3 - 7}{4}

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

x = arccos (\frac{3 - 7}{4}) + 2 πn, x = - arccos (\frac{3 - 7}{4}) + 2 πn

x = arccos (\frac{3 - 7}{4}) + 2 πn, x = - arccos (\frac{3 - 7}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{3 - 7}{4}) + 2 πn, x = - arccos (\frac{3 - 7}{4}) + 2 πn

Show solutions in decimal form

x = 1.80125 \dots + 2 πn, x = - 1.80125 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-2sqrt(3)cos(x)=sin^2(x) ?
The general solution for 3cos^2(x)-2sqrt(3)cos(x)=sin^2(x) is x=1.80125…+2pin,x=-1.80125…+2pin