Trig Functions- Could anyone tell me why my working is wrong?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
ku1005
Messages
66
Reaction score
0
Trig Functions- Could anyone tell me why my working is wrong??

the question seems simple..although i don't know what i am doing because it is obviously wrong!. thnkas heaps if anyone can help..

solve
2cosx-(sqrt(3))cosx sinx = 0 for (-pie<=x<=pie)

what i was doing is dividing through by cos x, therefore i obtain:

2-(sqrt(3))sinx= 0

however this statement is flawed! since no value exists for sin x= 2/(sqrt(3))

?it seems simple.. so wat am i doing wrong!1
 
Physics news on Phys.org
ku1005 said:
the question seems simple..although i don't know what i am doing because it is obviously wrong!. thnkas heaps if anyone can help..

solve
2cosx-(sqrt(3))cosx sinx = 0 for (-pie<=x<=pie)

what i was doing is dividing through by cos x, therefore i obtain:

2-(sqrt(3))sinx= 0

however this statement is flawed! since no value exists for sin x= 2/(sqrt(3))

?it seems simple.. so wat am i doing wrong!1

if you copied the question right then yes sin x = 2/srqt3 is not possible

however you CAN find answer for the cosine part
cos x = 0 what values of x will do?
 
yep got it now...instead i just did the follwing
=
cosx(2-(sqrt(3))sinx)=0
therfore either cosx= 0 (as u stated above) or the other applies...which isnlt possible...so yeah obviously the answers are pie/2 and -pie/2
thanks for your help!