:) If anyone can help me ? (C3)

If anyone can help me on these questions i would be eternally grateful.

1) Solve : 2(1-cos2x) = tan x

i did: 2(1-(1-2sin^2 x) = 4sin^2 x = tan x
4sin^2 x = sinx/cosx
4sinxcosx=0
2sin2x=0.

How do i solve it from here?

2)The curve with equation y=ln3x crosses x axis at the point P (p,0).
a)Sketch the graph of y=ln3x, showing the exact value of p.
I did this part. I basically drew the y=lnx, and changed the intercept from the (1,0) to (1/3,0), therefore p is 1/3.

But, I'm not sure how to do b)

b)Show that x=q is a solution of the equation x^2 + ln3x =0.
c)Show that x^2 + ln3x =0 can be rearragned to x=1/3e^(-x^2).

3)f=3(2^-x) -1
g=log2x

Find fg(x).

I got 3(2^-log2x) -1, but i don't know what to do from here.

I really appreciate any help. I'll make a mission of it to thank you all individually.

Thanks again

EDIT: I have now done Q2.

Reply 1

yusufu

1) Solve : 2(1-cos2x) = tan x

i did: 2(1-(1-2sin^2 x) = 4sin^2 x = tan x
4sin^2 x = sinx/cosx
4sinxcosx=0
2sin2x=0.

How do i solve it from here?

in this one you shouldnt divide through by sinx as you will "lose" solutions:
4sin^2 x = sinx/cosx
4sin^2x cosx = sinx
4sin^2x cosx - sinx = 0
sinx (4sinx cosx - 1) = 0
sinx(2sin2x-1)=0
sinx = 0, sin2x = 1/2
solve for x.

2)The curve with equation y=ln3x crosses x axis at the point P (p,0).
a)Sketch the graph of y=ln3x, showing the exact value of p.
I did this part. I basically drew the y=lnx, and changed the intercept from the (1,0) to (1/3,0), therefore p is 1/3.

But, I'm not sure how to do b)

b)Show that x=q is a solution of the equation x^2 + ln3x =0.

what is q? have you given all the info?

c)Show that x^2 + ln3x =0 can be rearragned to x=1/3e^(-x^2).

x^2 + ln3x =0
ln3x = -x^2
3x = e^(-x^2)
x= 1/3 e ^(-x^2)

3)f=3(2^-x) -1
g=log2x

Find fg(x).

I got 3(2^-log2x) -1, but i don't know what to do from here.

i think thats fine. nothing else to do. :smile: