Solve (cot^2 x)/(1 + cot^2 x) = 2sin2x

[Monkeyseat]

Hi,

Question

a) Show that (cot[sup:3dnj922s]2[/sup:3dnj922s]x)/(1 + cot[sup:3dnj922s]2[/sup:3dnj922s]x) = cos[sup:3dnj922s]2[/sup:3dnj922s]x

b) Hence solve (cot[sup:3dnj922s]2[/sup:3dnj922s]x)/(1 + cot[sup:3dnj922s]2[/sup:3dnj922s]x) = 2sin2x for 0 < x < 2pi, giving your answers in radians to three significant figures.

Working

a) I could do this part no problem.

b) I can't seem to get this in a 'solvable' format.

cos[sup:3dnj922s]2[/sup:3dnj922s]x = 2sin2x
cos[sup:3dnj922s]2[/sup:3dnj922s]x = 4sinxcosx

I don't know where to go from there... I tried writing cos[sup:3dnj922s]2[/sup:3dnj922s]x in terms of sinx but I couldn't get anywhere with it.

Any help with part (b) would be greatly appreciated.

Many thanks.

 

[galactus]

You've got it for the most part.

\(\displaystyle cos^{2}(x)=4sin(x)cos(x)\)

Assuming \(\displaystyle cos(x)\neq 0\)

\(\displaystyle cos(x)=4sin(x)\)

\(\displaystyle 4=\frac{cos(x)}{sin(x)}=cot(x)\)

\(\displaystyle x=tan^{-1}(\frac{1}{4})+C{\pi} \;\ or \;\ x=\frac{(2C-1){\pi}}{2}\)

You should have 4 solutions in the given interval.

 

[Monkeyseat]

You've got it for the most part.

\(\displaystyle cos^{2}(x)=4sin(x)cos(x)\)

Assuming \(\displaystyle cos(x)\neq 0\)

\(\displaystyle cos(x)=4sin(x)\)

\(\displaystyle 4=\frac{cos(x)}{sin(x)}=cot(x)\)

\(\displaystyle x=tan^{-1}(\frac{1}{4})+C{\pi} \;\ or \;\ x=\frac{(2C-1){\pi}}{2}\)

You should have 4 solutions in the given interval.

Thanks for the reply. The method I have for doing these questions has only yielded two solutions.

I got to tan x = 1/4.

So the principal value is tan^-1 (1/4) which is 0.245.

Using a CAST diagram, I got 0.245 and 3.39 as my solutions. I am obviously missing 2 solutions. How can I get the other 2 (using this method if possible, I have got used to doing them this way now - I didn't understand what you had at the end of your reply).

When I was looking at this example in the textbook:

2sinxcosx = cosx

It said "do not cancel the cosx terms". I always thought cancelling the cosx term would lose solutions (as seems to be the case for this question using my method).

Thanks.

 

[Deleted member 4993]

The other two solutions are

x = ?/2 and x = 3?/2 where cos(x) = 0 [ those are the solutions you loose if you simply "cancel" out cos(x)]

In your graphing calculator plot

y = cos[sup:3rvn4bjp]2[/sup:3rvn4bjp](x) - 2sin(2x)

and overlay

y = cos(x) - 4sin(x)

then compare the roots.