#1

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

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

tan 2x(4cosx+2)=0

#2

is it exact values? (calc or non-calc)

#3

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)

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

or

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

#4

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

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

#5

Do pay attention to that post - the equation in the first post is ambiguous. Which is it meant to be?dont pay attention to the post above

#6

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

#7

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)

2x=180.n(nez,n can be any integer)

x=90.n

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

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,*