# Two sin equations

• Dec 17th 2010, 06:09 AM
klik11
Two sinus equations
The first equation is

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

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

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

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

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

Thank you
• Dec 17th 2010, 06:29 AM
Prove It
You have gotten to $\displaystyle \displaystyle \sin{x} = -\frac{1}{15}$. You need to remember that sine is negative in the third and fourth quadrants.

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

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

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

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

$\displaystyle \displaystyle x = 30^{\circ} + 120^{\circ}K$.
• Dec 21st 2010, 03:37 PM
Quote:

Originally Posted by klik11
The first equation is

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

I tried solving it:

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

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

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

But the book says it's

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

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

Thank you

The book answers are the solution to

$\displaystyle 7sinx+5=0$