What is the general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi ?

Q: What is the general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi ?

The general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi is x=2.15146…,x=-2.15146…+2pi

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Popular >

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

Solution

3 cos^{2} (x) - 2 cos (x) - 2 = 0, 0 \leq x < 2 π

Solution

x = 2.15146 \dots, x = - 2.15146 \dots + 2 π

Degrees

x = 123.26990 \dots^{\circ}, x = 236.73009 \dots^{\circ}

Solution steps

3 cos^{2} (x) - 2 cos (x) - 2 = 0, 0 \leq x < 2 π

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 3 \cdot 2 = 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:

n ab = n a n b

= 7 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 27

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

Separate the solutions

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

u = \frac{- ( - 2 ) + 2 7}{2 \cdot 3} : \frac{1 + 7}{3}

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2 + 2 7}{6}

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 )}{6}

Cancel the common factor:

2

= \frac{1 + 7}{3}

u = \frac{- ( - 2 ) - 2 7}{2 \cdot 3} : \frac{1 - 7}{3}

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2 - 2 7}{6}

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 )}{6}

Cancel the common factor:

2

= \frac{1 - 7}{3}

The solutions to the quadratic equation are:

u = \frac{1 + 7}{3}, u = \frac{1 - 7}{3}

Substitute back

u = cos (x)

cos (x) = \frac{1 + 7}{3}, cos (x) = \frac{1 - 7}{3}

cos (x) = \frac{1 + 7}{3}, cos (x) = \frac{1 - 7}{3}

cos (x) = \frac{1 + 7}{3}, 0 \leq x < 2 π :

No Solution

cos (x) = \frac{1 + 7}{3}, 0 \leq x < 2 π

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1 - 7}{3}, 0 \leq x < 2 π : x = arccos (\frac{1 - 7}{3}), x = - arccos (\frac{1 - 7}{3}) + 2 π

cos (x) = \frac{1 - 7}{3}, 0 \leq x < 2 π

Apply trig inverse properties

cos (x) = \frac{1 - 7}{3}

General solutions for

cos (x) = \frac{1 - 7}{3}

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

x = arccos (\frac{1 - 7}{3}) + 2 πn, x = - arccos (\frac{1 - 7}{3}) + 2 πn

x = arccos (\frac{1 - 7}{3}) + 2 πn, x = - arccos (\frac{1 - 7}{3}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = arccos (\frac{1 - 7}{3}), x = - arccos (\frac{1 - 7}{3}) + 2 π

Combine all the solutions

x = arccos (\frac{1 - 7}{3}), x = - arccos (\frac{1 - 7}{3}) + 2 π

Show solutions in decimal form

x = 2.15146 \dots, x = - 2.15146 \dots + 2 π

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi ?
The general solution for 3cos^2(x)-2cos(x)-2=0,0<= x<2pi is x=2.15146…,x=-2.15146…+2pi