What is the general solution for cos(2θ)=12(1+cos(2θ)) ?

The general solution for cos(2θ)=12(1+cos(2θ)) is No Solution for θ\in\mathbb{R}

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cos(2θ)=12(1+cos(2θ))

cos (2 θ) = 12 (1 + cos (2 θ))

No Solution for θ \in R

Solution steps

cos (2 θ) = 12 (1 + cos (2 θ))

Solve by substitution

cos (2 θ) = 12 (1 + cos (2 θ))

Let:

cos (2 θ) = u

u = 12 (1 + u)

u = 12 (1 + u) : u = - \frac{12}{11}

u = 12 (1 + u)

Expand

12 (1 + u) : 12 + 12 u

12 (1 + u)

Apply the distributive law:

a (b + c) = ab + a c

a = 12, b = 1, c = u

= 12 \cdot 1 + 12 u

Multiply the numbers:

12 \cdot 1 = 12

= 12 + 12 u

u = 12 + 12 u

Move

12 u

to the left side

u = 12 + 12 u

Subtract

12 u

from both sides

u - 12 u = 12 + 12 u - 12 u

Simplify

- 11 u = 12

- 11 u = 12

Divide both sides by

- 11

- 11 u = 12

Divide both sides by

- 11

\frac{- 11 u}{- 11} = \frac{12}{- 11}

Simplify

u = - \frac{12}{11}

u = - \frac{12}{11}

Substitute back

u = cos (2 θ)

cos (2 θ) = - \frac{12}{11}

cos (2 θ) = - \frac{12}{11}

cos (2 θ) = - \frac{12}{11} :

No Solution

cos (2 θ) = - \frac{12}{11}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for θ \in R

0 = e^{t a n (x)} - 2

cot (x) = \frac{3}{2}

4 tan^{2} (x) - 1 = 0

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

2 sin (θ) cos (θ) = cos (θ)

What is the general solution for cos(2θ)=12(1+cos(2θ)) ?
The general solution for cos(2θ)=12(1+cos(2θ)) is No Solution for θ\in\mathbb{R}