Trig equations

**classicstrings** · July 20th 2006, 08:21 AM

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

**galactus** · July 20th 2006, 08:47 AM

Solve for x:

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

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

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

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

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

**ThePerfectHacker** · July 20th 2006, 09:39 AM

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,
$\sin x =-\frac{1}{2}$
Then, then the two coterminal angles are,
$x=\frac{4\pi}{3}$
$x=\frac{5\pi}{3}$
Then all solutions are,
$x=\frac{4\pi}{3}+2\pi k$
$x=\frac{5\pi}{3}+2\pi k$
But since you have,
$\sin \left( \frac{\pi x}{12} \right) =-\frac{1}{2}$
All solutions are,
$\frac{\pi x}{12}=\frac{4\pi}{3}+2\pi k$
$\frac{\pi x}{12}=\frac{5\pi}{3}+2\pi k$
Thus, (mutiply by reciprocal)
$x=16+24k$
$x=20+24k$

**CaptainBlack** · July 20th 2006, 09:53 AM

Originally Posted by ThePerfectHacker

$x=16+24k$
$x=20+24k$

$k=0,\ \pm 1,\ \pm 2, ..$

**classicstrings** · July 25th 2006, 02:00 AM

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

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