What is the general solution for 2+cos(2θ)=0 ?

The general solution for 2+cos(2θ)=0 is No Solution for θ\in\mathbb{R}

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2+cos(2θ)=0

2 + cos (2 θ) = 0

No Solution for θ \in R

Solution steps

2 + cos (2 θ) = 0

Move

2

to the right side

2 + cos (2 θ) = 0

Subtract

2

from both sides

2 + cos (2 θ) - 2 = 0 - 2

Simplify

cos (2 θ) = - 2

cos (2 θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution for θ \in R

cos (\frac{π ( x + 5 )}{3}) = \frac{1}{2}

(sin (x))^{2} + (1 - 2) sin (x) - 2 = 0

4 sin^{2} (A) - 1 = 0

tan (θ) = \frac{22}{7}

u, x = 4 tanh (u)

What is the general solution for 2+cos(2θ)=0 ?
The general solution for 2+cos(2θ)=0 is No Solution for θ\in\mathbb{R}