What is the general solution for 1+cos(x)=sqrt(3)cos(x) ?

The general solution for 1+cos(x)=sqrt(3)cos(x) is No Solution for x\in\mathbb{R}

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Popular >

1+cos(x)=sqrt(3)cos(x)

Solution

1 + cos (x) = 3 cos (x)

Solution

No Solution for x \in R

Solution steps

1 + cos (x) = 3 cos (x)

Solve by substitution

1 + cos (x) = 3 cos (x)

Let:

cos (x) = u

1 + u = 3 u

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

1 + u = 3 u

Move

1

to the right side

1 + u = 3 u

Subtract

1

from both sides

1 + u - 1 = 3 u - 1

Simplify

u = 3 u - 1

u = 3 u - 1

Move

3 u

to the left side

u = 3 u - 1

Subtract

3 u

from both sides

u - 3 u = 3 u - 1 - 3 u

Simplify

u - 3 u = - 1

u - 3 u = - 1

Factor

u - 3 u : (1 - 3) u

u - 3 u

Factor out common term

u

= u (1 - 3)

(1 - 3) u = - 1

Divide both sides by

1 - 3

(1 - 3) u = - 1

Divide both sides by

1 - 3

\frac{( 1 - 3 ) u}{1 - 3} = \frac{- 1}{1 - 3}

Simplify

\frac{( 1 - 3 ) u}{1 - 3} = \frac{- 1}{1 - 3}

Simplify

\frac{( 1 - 3 ) u}{1 - 3} : u

\frac{( 1 - 3 ) u}{1 - 3}

Cancel the common factor:

1 - 3

= u

Simplify

\frac{- 1}{1 - 3} : \frac{3 + 1}{2}

\frac{- 1}{1 - 3}

Apply the fraction rule:

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

1 - 3 = - (3 - 1)

= \frac{1}{3 - 1}

Rationalize

\frac{1}{3 - 1} : \frac{1 + 3}{2}

\frac{1}{3 - 1}

Multiply by the conjugate

\frac{3 + 1}{3 + 1}

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

1 \cdot (3 + 1) = 3 + 1

(3 - 1) (3 + 1) = 2

(3 - 1) (3 + 1)

Apply Difference of Two Squares Formula:

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

a = 3, b = 1

= (3)^{2} - 1^{2}

Simplify

(3)^{2} - 1^{2} : 2

(3)^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} - 1

(3)^{2} = 3

(3)^{2}

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

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

= \frac{3 + 1}{2}

= \frac{3 + 1}{2}

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

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

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

Substitute back

u = cos (x)

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

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

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 1+cos(x)=sqrt(3)cos(x) ?
The general solution for 1+cos(x)=sqrt(3)cos(x) is No Solution for x\in\mathbb{R}