Two sin equations

**klik11** · December 17th 2010, 06:09 AM

The first equation is

$75*\sin{x}+5=0$

I tried solving it:
$\sin{x}=\frac{-5}{75}$
$x1 = -3.82^{\circ} + 360K$
$x2 = 183.82^{\circ} + 360K$

But the book says it's
$x1 = -45.58^{\circ} + 360K$
$x2 = 225.58^{\circ} + 360K$

The second equation looks more simple and I can solve it with my calculator but I want to know how to solve it without my calculator

$\sin{3x} = 1$

Thank you

**Prove It** · December 17th 2010, 06:29 AM

You have gotten to $\displaystyle \sin{x} = -\frac{1}{15}$ . You need to remember that sine is negative in the third and fourth quadrants.

So your solutions are $\displaystyle \left\{ 180^{\circ} + \sin^{-1}{\left(\frac{1}{15}\right)}, 360^{\circ} - \sin^{-1}{\left(\frac{1}{15}\right)}\right\} + 360^{\circ}K$ .

For the second, $\displaystyle \sin{X} = 1$ if $\displaystyle X = 90^{\circ} + 360^{\circ}K$ .

So if $\displaystyle \sin{3x} = 1$

$\displaystyle 3x = 90^{\circ} + 360^{\circ}K$

$\displaystyle x = 30^{\circ} + 120^{\circ}K$ .

**Archie Meade** · December 21st 2010, 03:37 PM

Originally Posted by klik11

The first equation is

$75*\sin{x}+5=0$

I tried solving it:

$\sin{x}=\frac{-5}{75}$

$x1 = -3.82^{\circ} + 360K$

$x2 = 183.82^{\circ} + 360K$

But the book says it's

$x1 = -45.58^{\circ} + 360K$

$x2 = 225.58^{\circ} + 360K$

Thank you

The book answers are the solution to

$7sinx+5=0$

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