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

The general solution for 3cos^2(x)-2sqrt(3)cos(x)=cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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Popular >

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

3 u^{2} - 23 u = u

3 u^{2} - 23 u = u : u = \frac{2 3 + 1}{3}, u = 0

3 u^{2} - 23 u = u

Move

u

to the left side

3 u^{2} - 23 u = u

Subtract

u

from both sides

3 u^{2} - 23 u - u = u - u

Simplify

3 u^{2} - 23 u - u = 0

3 u^{2} - 23 u - u = 0

Write in the standard form

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

3 u^{2} - (23 + 1) u = 0

Solve with the quadratic formula

3 u^{2} - (23 + 1) u = 0

Quadratic Equation Formula:

For

a = 3, b = - 23 - 1, c = 0

u_{1, 2} = \frac{- ( - 2 3 - 1 ) \pm ( - 2 3 - 1 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 2 3 - 1 ) \pm ( - 2 3 - 1 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 23 - 1)^{2} - 4 \cdot 3 \cdot 0 = 23 + 1

(- 23 - 1)^{2} - 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= (- 23 - 1)^{2} - 0

(- 23 - 1)^{2} - 0 = (- 23 - 1)^{2}

= (- 23 - 1)^{2}

Expand

(- 23 - 1)^{2} : 13 + 43

(- 23 - 1)^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = - 23, b = 1

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

Simplify

(- 23)^{2} - 2 (- 23) \cdot 1 + 1^{2} : 13 + 43

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Apply rule

- (- a) = a

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

(- 23)^{2} = 12

(- 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

2^{2} = 4

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

2 \cdot 23 \cdot 1 = 43

2 \cdot 23 \cdot 1

Multiply the numbers:

2 \cdot 2 \cdot 1 = 4

= 43

= 12 + 43 + 1

Add the numbers:

12 + 1 = 13

= 13 + 43

= 13 + 43

= 13 + 43

= 12 + 43 + 1

= 4 \cdot 3 + 43 + 1

= (4)^{2} (3)^{2} + 43 + (1)^{2}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

1 = 1

1

Apply rule

1 = 1

= 1

= 2^{2} (3)^{2} + 43 + 1^{2}

2 \cdot 23 \cdot 1 = 43

2 \cdot 23 \cdot 1

Multiply the numbers:

2 \cdot 2 \cdot 1 = 4

= 43

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

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

= (23 + 1)^{2}

Apply radical rule:

(23 + 1)^{2} = 23 + 1

= 23 + 1

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

Separate the solutions

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

u = \frac{- ( - 2 3 - 1 ) + 2 3 + 1}{2 \cdot 3} : \frac{2 3 + 1}{3}

\frac{- ( - 2 3 - 1 ) + 2 3 + 1}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- ( - 2 3 - 1 ) + 2 3 + 1}{6}

Expand

- (- 23 - 1) + 23 + 1 : 43 + 2

- (- 23 - 1) + 23 + 1

- (- 23 - 1) : 23 + 1

- (- 23 - 1)

Distribute parentheses

= - (- 23) - (- 1)

Apply minus-plus rules

- (- a) = a

= 23 + 1

= 23 + 1 + 23 + 1

Simplify

23 + 1 + 23 + 1 : 43 + 2

23 + 1 + 23 + 1

Add similar elements:

23 + 23 = 43

= 43 + 1 + 1

Add the numbers:

1 + 1 = 2

= 43 + 2

= 43 + 2

= \frac{4 3 + 2}{6}

Factor

43 + 2 : 2 (23 + 1)

43 + 2

Rewrite as

= 2 \cdot 23 + 2 \cdot 1

Factor out common term

2

= 2 (23 + 1)

= \frac{2 ( 2 3 + 1 )}{6}

Cancel the common factor:

2

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

u = \frac{- ( - 2 3 - 1 ) - ( 2 3 + 1 )}{2 \cdot 3} : 0

\frac{- ( - 2 3 - 1 ) - ( 2 3 + 1 )}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- ( - 2 3 - 1 ) - ( 2 3 + 1 )}{6}

Expand

- (- 23 - 1) - (23 + 1) : 0

- (- 23 - 1) - (23 + 1)

- (- 23 - 1) : 23 + 1

- (- 23 - 1)

Distribute parentheses

= - (- 23) - (- 1)

Apply minus-plus rules

- (- a) = a

= 23 + 1

= 23 + 1 - (23 + 1)

- (23 + 1) : - 23 - 1

- (23 + 1)

Distribute parentheses

= - (23) - (1)

Apply minus-plus rules

+ (- a) = - a

= - 23 - 1

= 23 + 1 - 23 - 1

Simplify

23 + 1 - 23 - 1 : 0

23 + 1 - 23 - 1

Add similar elements:

23 - 23 = 0

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

= 0

= \frac{0}{6}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = \frac{2 3 + 1}{3}, u = 0

Substitute back

u = cos (x)

cos (x) = \frac{2 3 + 1}{3}, cos (x) = 0

cos (x) = \frac{2 3 + 1}{3}, cos (x) = 0

cos (x) = \frac{2 3 + 1}{3} :

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

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

cos (x) = 0

General solutions for

cos (x) = 0

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}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

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196 sin (θ) - 49 cos (θ) = 160

4tan(x)+3=-7

4 tan (x) + 3 = - 7

8tan(θ)+15=0

8 tan (θ) + 15 = 0

cos(x)=(48.4)/(54.5)

cos (x) = \frac{48.4}{54.5}

(sin(x^3+2x-1))a=1

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

Frequently Asked Questions (FAQ)

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