What is the general solution for 4cos^2(x)+4cos(x)=1 ?

The general solution for 4cos^2(x)+4cos(x)=1 is x=1.36217…+2pin,x=2pi-1.36217…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4cos^2(x)+4cos(x)=1

Solution

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

Solution

x = 1.36217 \dots + 2 πn, x = 2 π - 1.36217 \dots + 2 πn

Degrees

x = 78.04714 \dots^{\circ} + 36 0^{\circ} n, x = 281.95285 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

4 u^{2} + 4 u = 1

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

4 u^{2} + 4 u = 1

Move

1

to the left side

4 u^{2} + 4 u = 1

Subtract

1

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

4^{2} - 4 \cdot 4 (- 1) = 42

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

Apply rule

- (- a) = a

= 4^{2} + 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} + 16

4^{2} = 16

= 16 + 16

Add the numbers:

16 + 16 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{4} \cdot 2

Apply radical rule:

= 2 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

Refine

= 42

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

Separate the solutions

u_{1} = \frac{- 4 + 4 2}{2 \cdot 4}, u_{2} = \frac{- 4 - 4 2}{2 \cdot 4}

u = \frac{- 4 + 4 2}{2 \cdot 4} : \frac{- 1 + 2}{2}

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 + 4 2}{8}

Factor

- 4 + 42 : 4 (- 1 + 2)

- 4 + 42

Rewrite as

= - 4 \cdot 1 + 42

Factor out common term

4

= 4 (- 1 + 2)

= \frac{4 ( - 1 + 2 )}{8}

Cancel the common factor:

4

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

u = \frac{- 4 - 4 2}{2 \cdot 4} : - \frac{1 + 2}{2}

\frac{- 4 - 4 2}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 - 4 2}{8}

Factor

- 4 - 42 : - 4 (1 + 2)

- 4 - 42

Rewrite as

= - 4 \cdot 1 - 42

Factor out common term

4

= - 4 (1 + 2)

= - \frac{4 ( 1 + 2 )}{8}

Cancel the common factor:

4

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

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

x = 1.36217 \dots + 2 πn, x = 2 π - 1.36217 \dots + 2 πn

Popular Examples

sin(x)= 1/6

sin (x) = \frac{1}{6}

tan(θ)= 2/5

tan (θ) = \frac{2}{5}

tan(x)+3cot(x)=4

tan (x) + 3 cot (x) = 4

cos(2x)[cos(2x)-3]=0

cos (2 x) [cos (2 x) - 3] = 0

cos^2(θ)= 1/4

cos^{2} (θ) = \frac{1}{4}

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)+4cos(x)=1 ?
The general solution for 4cos^2(x)+4cos(x)=1 is x=1.36217…+2pin,x=2pi-1.36217…+2pin