What is the general solution for 2cos(x)+1=0-2,pi<= x<= 2pi ?

The general solution for 2cos(x)+1=0-2,pi<= x<= 2pi is No Solution for x\in\mathbb{R}

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2cos(x)+1=0-2,pi<= x<= 2pi

2 cos (x) + 1 = 0 - 2, π \leq x \leq 2 π

No Solution for x \in R

Solution steps

2 cos (x) + 1 = 0 - 2, π \leq x \leq 2 π

Subtract the numbers:

0 - 2 = - 2

2 cos (x) + 1 = - 2

Move

1

to the right side

2 cos (x) + 1 = - 2

Subtract

1

from both sides

2 cos (x) + 1 - 1 = - 2 - 1

Simplify

2 cos (x) = - 3

2 cos (x) = - 3

Divide both sides by

2

2 cos (x) = - 3

Divide both sides by

2

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

Simplify

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

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

- 1 \leq cos (x) \leq 1

No Solution for x \in R

cos^{3} (x) = - cos (x)

csc (2 x) = 2 sin (2 x), 0 \leq x \leq 2 π

cos (x) = - \frac{12}{13}

x, z = arcsin (\frac{x}{y})

7 tan^{2} (θ) - 11 tan (θ) - 4 = - 8 tan (θ)

What is the general solution for 2cos(x)+1=0-2,pi<= x<= 2pi ?
The general solution for 2cos(x)+1=0-2,pi<= x<= 2pi is No Solution for x\in\mathbb{R}