I need some help on this. any help would be much appericated. thanks

Solve for 0 < less than or equal to 0 < 2Pi

tan 2x(4cosx+2)=0
is it exact values? (calc or non-calc)
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could you show function grouping a bit better? I'm not sure if you mean

tan(2x)*4cos(x+2)

or

tan(2x)*(4cos(x)+2)
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ok useing zero product property:
set tan2x=0 and 4cosx+2=0
then use double angle formula to solve the tan part and subtract 2 devide by four to solve the other part (cosx=-2/4 or of course -1/2)

dont pay attention to the post above
If its not fender than whatever]
Quote by ffud
dont pay attention to the post above
Do pay attention to that post - the equation in the first post is ambiguous. Which is it meant to be?
Quote by ffud
ok useing zero product property:
then use double angle formula to solve the tan part
dont pay attention to the post above

You don't even need to do that. If tan2x = 0, then 2x = 0, 2pi, 4pi
Therefore x = 0, pi, 2pi

Yea do that for cos ie cosx = -1/2, then x = 2pi/3, 4pi/3
Quote by mustardman
Granted, they're crap, but they're decent.
Firstly use the property of zero i.e)
tan2x(4cosx+2)=0
so:
tan2x=0 or cosx=-0.5(0,5 is one of the ratios of your special triangles,so you know when you see this the ratio will be a 30 ref angle)
therefore according to general formula:
tan2x is positive and negative in quads 1 and 2(only use quads 1 and 2 because of the period of the tan graph)

q1
2x=180.n(nez,n can be any integer)
x=90.n
q2
2x=180-0+180.n(ref angle of tan 2x=zero
x=90+90n
and likewise for cos:
gives you:
x=150+360n(q2)
x=210+360n(q3)
if 2pi>0 then pi<0(if you divide a inequality the sign changes)
therefore
Pi<0<x<infinity
smallest value for x=0,therefore XE(0;+infinity)...( Pi has to be less than 0 so zero can be included)
please fell free to pm me if they are any questions,this question will probably be about 10 marks in a exam
Last edited by u-legend at May 19, 2008,