# inverse trigonometric functions

• Nov 28th 2009, 05:47 PM
zukywich08
inverse trigonometric functions
just need some help getting started

1.) sin2x= -1/2
2.) 4cos^2 x-8cosx+1=0
3.) sin2x= -sinx
greatly appreciate it(Nod)
• Nov 28th 2009, 06:10 PM
skeeter
Quote:

Originally Posted by zukywich08
just need some help getting started

1.) sin2x= -1/2
2.) 4cos^2 x-8cosx+1=0
3.) sin2x= -sinx
greatly appreciate it(Nod)

(1) $sin(u) = -\frac{1}{2}$

$u = \frac{7\pi}{6} \, , \, \frac{11\pi}{6}$ and those angles coterminal with these two.

(3) $2\sin{x}\cos{x} + \sin{x} = 0$

factor and solve
• Nov 28th 2009, 06:31 PM
zukywich08
quick question
for the third problem how would you factor it? do you take sinx out of it?
• Nov 28th 2009, 06:34 PM
skeeter
Quote:

Originally Posted by zukywich08
for the third problem how would you factor it? do you take sinx out of it?

yes
• Nov 28th 2009, 06:39 PM
zukywich08
i got sinx(sinxcosx+1)=0
but i dont know what to do with the cosine?
• Nov 28th 2009, 06:41 PM
skeeter
Quote:

Originally Posted by zukywich08
i got sinx(sinxcosx+1)=0
but i dont know what to do with the cosine?

$2\sin{x}\cos{x} + \sin{x} = 0$
$\sin{x}(2\cos{x} + 1) = 0$