# Trigonometric function

• Nov 18th 2008, 01:17 PM
cjru
Trigonometric function
Hello guys!

I have a problem solving this equation. I really dont know how to start Could someone pls help me with this...

The problem is that Find all solutions of the given equation in the indicated interval.

3sin2X+cosX = 0 [-pi/2, pi/2]

• Nov 18th 2008, 01:32 PM
flyingsquirrel
Hi,

Hint : $\displaystyle \sin (2x)=2\sin x\cos x$.
• Nov 18th 2008, 01:56 PM
cjru
thank you very much for replying.

OK so using that identity my equation

from 3sin2X+cox

will become (3)(2)sinxcosx + cosx = 0

6sinxcosx + cox = 0

Im I doing it right?
• Nov 18th 2008, 02:01 PM
11rdc11
Quote:

Originally Posted by cjru
thank you very much for replying.

OK so using that identity my equation

from 3sin2X+cox

will become (3)(2)sinxcosx + cosx = 0

6sinxcosx + cox = 0

Im I doing it right?

Yes then factor out a cos x
• Nov 18th 2008, 02:13 PM
cjru
Quote:

Originally Posted by 11rdc11
Yes then factor out a cos x

ok then I get cosX (6sinX) = 0

Therefore Cos X = 0 where in Cos can only be zero at x equal to pi/2 and -pi/2

Is this right?

How about the 6sinx = 0

Pls check if i did it right....

sin x = 0/6 then sin x = 0

Therefore sin x = 0 wherein sin can only be zero at x equal to 0 and pi.
• Nov 18th 2008, 02:39 PM
skeeter
not quite ...

$\displaystyle \cos{x}(6\sin{x} + 1) = 0$
• Nov 18th 2008, 05:53 PM
cjru
o yeah! I over look it. So what will be the solution for 6sinX+1?

If I equate that to zero sinx = 1/6 Is this correct?(Worried)
• Nov 18th 2008, 08:20 PM
11rdc11
~
Quote:

Originally Posted by cjru
o yeah! I over look it. So what will be the solution for 6sinX+1?

If I equate that to zero sinx = 1/6 Is this correct?(Worried)

$\displaystyle \sin{x} = -~\frac{1}{6}$

$\displaystyle x = \sin^{-1} {\bigg(-~\frac{1}{6}\bigg)}$
• Nov 18th 2008, 09:32 PM
cjru
Thank you so much. Math help forum is the best. :)

God bless and more power to you guys!

Again thank you(Clapping)