What is the general solution for sqrt(1-cos(3x))=2pi ?

The general solution for sqrt(1-cos(3x))=2pi is No Solution for x\in\mathbb{R}

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sqrt(1-cos(3x))=2pi

1 - cos (3 x) = 2 π

No Solution for x \in R

Solution steps

1 - cos (3 x) = 2 π

Square both sides:

1 - cos (3 x) = 4 π^{2}

1 - cos (3 x) = 4 π^{2}

Solve

1 - cos (3 x) = 4 π^{2} : cos (3 x) = - 4 π^{2} + 1

cos (3 x) = - 4 π^{2} + 1

Verify Solutions:

cos (3 x) = - 4 π^{2} + 1

True

The solution is

cos (3 x) = - 4 π^{2} + 1

- 1 \leq cos (x) \leq 1

No Solution for x \in R

(sin (x - 4)) (sin (x - 1)) = 0

θ, r = \frac{10}{3 + 2 cos ( θ )}

sin (x) = \frac{17.75}{42}

tan (θ) - sec (θ) = 3, 0 \leq θ < 2 π

tan (θ) = \frac{5}{2}, 0 < θ < \frac{π}{2}, sin (\frac{θ}{2})

What is the general solution for sqrt(1-cos(3x))=2pi ?
The general solution for sqrt(1-cos(3x))=2pi is No Solution for x\in\mathbb{R}