Solve this equation????

[zibb3r]

Solve sin(2x) + 2sin(2x)cos(2x) = 0 where x is [-pi,3pi]


Not sure, any idea how?

THanks

[xZero]

, where

solve for 2x within the restriction and divide both side by 2 to find x

[zibb3r]

Would you e able to explain further?? Im still a bit confused

[taiga]

You can then break


into

and

From there you should be able to solve it

If you can't (I'm not saying this in a mean way) then you should check out some of your earlier maths books

[xZero]

^this, looking through books and working it out yourself will be more beneficial than us posting the solution. Of course if you can do it then ignore this

[zibb3r]

I can do that and the null factor law just confused about what to do next.......

My text book is at school aswell.....

[xZero]

try replacing 2x with u and solve for u, in this case it will be sin(u)=0 or 1+2cos(u)=0. (remember the domain of 2x)

[zibb3r]

Nevermind, thanks to all.  I guess trig just isnt my thing! Will leave it:D

[brightsky]

sin(2x) + 2 sin(2x) cos(2x) = 0
sin(2x) (1 + 2 cos(2x)) = 0
sin(2x) = 0 or 1 + 2cos(2x) = 0
from sin(2x) = 0, we get:
2x = -2pi, -pi, 0, pi, 2pi, 3pi, 4pi, 5pi, 6pi [remember restriction]
x = -pi, -pi/2, 0, pi/2, pi, 3pi/2, 2pi, 5pi/2, 3pi
from 1 + 2cos(2x) = 0
we get cos(2x) = -1/2
so:
2x = -4pi/3, -2pi/3, 2pi/3, 4pi/3, 8pi/3, 10pi/3, 14pi/3, 16pi/3
x = -2pi/3, -pi/3, pi/3, 2pi/3, 4pi/3, 5pi/3, 7pi/3, 8pi/3
so we have a total of...17 solutions...