# Math Help - inverse trigonometric functions

1. ## 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

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

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

(2) quadratic formula ...

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

factor and solve

3. ## quick question

for the third problem how would you factor it? do you take sinx out of it?

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

5. i got sinx(sinxcosx+1)=0
but i dont know what to do with the cosine?

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

$2\sin{x}\cos{x} + \sin{x} = 0$

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

set each factor = 0 and solve

7. alright thanks for the help man it came in handy