What is the general solution for 2sin(4x)+8=0 ?

The general solution for 2sin(4x)+8=0 is No Solution for x\in\mathbb{R}

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2sin(4x)+8=0

2 sin (4 x) + 8 = 0

No Solution for x \in R

Solution steps

2 sin (4 x) + 8 = 0

Move

8

to the right side

2 sin (4 x) + 8 = 0

Subtract

8

from both sides

2 sin (4 x) + 8 - 8 = 0 - 8

Simplify

2 sin (4 x) = - 8

2 sin (4 x) = - 8

Divide both sides by

2

2 sin (4 x) = - 8

Divide both sides by

2

\frac{2 sin ( 4 x )}{2} = \frac{- 8}{2}

Simplify

\frac{2 sin ( 4 x )}{2} = \frac{- 8}{2}

Simplify

\frac{2 sin ( 4 x )}{2} : sin (4 x)

\frac{2 sin ( 4 x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin (4 x)

Simplify

\frac{- 8}{2} : - 4

\frac{- 8}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= - 4

sin (4 x) = - 4

sin (4 x) = - 4

sin (4 x) = - 4

- 1 \leq sin (x) \leq 1

No Solution for x \in R

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2 csc (x) + 2 = 0

1 + cos (x) = 3 cos (x)

cos (θ) = \frac{14}{16}

2 sec (x^{2}) + tan (x^{2}) - 3 = 0

What is the general solution for 2sin(4x)+8=0 ?
The general solution for 2sin(4x)+8=0 is No Solution for x\in\mathbb{R}