What is the general solution for 5sec(2x)+2=0 ?

The general solution for 5sec(2x)+2=0 is No Solution for x\in\mathbb{R}

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5sec(2x)+2=0

5 sec (2 x) + 2 = 0

No Solution for x \in R

Solution steps

5 sec (2 x) + 2 = 0

Move

2

to the right side

5 sec (2 x) + 2 = 0

Subtract

2

from both sides

5 sec (2 x) + 2 - 2 = 0 - 2

Simplify

5 sec (2 x) = - 2

5 sec (2 x) = - 2

Divide both sides by

5

5 sec (2 x) = - 2

Divide both sides by

5

\frac{5 sec ( 2 x )}{5} = \frac{- 2}{5}

Simplify

sec (2 x) = - \frac{2}{5}

sec (2 x) = - \frac{2}{5}

sec (x) \leq - 1 or sec (x) \geq 1

No Solution for x \in R

2.1 sin (θ) = 1.33 sin (θ + 9 0^{\circ})

1 sin (4 5^{\circ}) = 1.33 sin (θ)

3 cot^{2} (x) = 9

1 2^{(2)} = 1 3^{(2)} + 5^{(2)} - 2 (13) (5) cos (B)

tan (6 t) = 2

What is the general solution for 5sec(2x)+2=0 ?
The general solution for 5sec(2x)+2=0 is No Solution for x\in\mathbb{R}