# Thread: Trig equations

1. ## Trig equations

How do I solve this?

-0.5 = sin((Pi*x)/12)

Sin is -0.5 when the angle is 30. Since it is negative, angles are 210 and 330.

I get eventually 7,11,14,23...

Which are wrong somehow. Please help! Thx

2. Solve for x:

$\displaystyle sin^{-1}(-1/2)=\frac{{\pi}x}{12}$

$\displaystyle x=\frac{12sin^{-1}(-1/2)}{\pi}=-2$

$\displaystyle \frac{{\pi}(-2)}{12}=\frac{-\pi}{6}$

$\displaystyle sin(\frac{-5\pi}{6}+2n{\pi})=-1/2$

$\displaystyle sin(\frac{-\pi}{6}+2n{\pi})=-1/2$, n=0,1,2,......

3. Originally Posted by classicstrings
How do I solve this?

-0.5 = sin((Pi*x)/12)

Sin is -0.5 when the angle is 30. Since it is negative, angles are 210 and 330.

I get eventually 7,11,14,23...

Which are wrong somehow. Please help! Thx
If you simply had,
$\displaystyle \sin x =-\frac{1}{2}$
Then, then the two coterminal angles are,
$\displaystyle x=\frac{4\pi}{3}$
$\displaystyle x=\frac{5\pi}{3}$
Then all solutions are,
$\displaystyle x=\frac{4\pi}{3}+2\pi k$
$\displaystyle x=\frac{5\pi}{3}+2\pi k$
But since you have,
$\displaystyle \sin \left( \frac{\pi x}{12} \right) =-\frac{1}{2}$
All solutions are,
$\displaystyle \frac{\pi x}{12}=\frac{4\pi}{3}+2\pi k$
$\displaystyle \frac{\pi x}{12}=\frac{5\pi}{3}+2\pi k$
Thus, (mutiply by reciprocal)
$\displaystyle x=16+24k$
$\displaystyle x=20+24k$

4. Originally Posted by ThePerfectHacker
$\displaystyle x=16+24k$
$\displaystyle x=20+24k$
$\displaystyle k=0,\ \pm 1,\ \pm 2, ..$

5. Thanks a lot guys! I realised today when I logged on that you changed the updated the number of (1800+) members sign too!