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1. Hence or otherwise show that the equation:

cosx (cos2x/cosx + sinx) = 1/2, can be written as:

sin2x = cos2x

afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
2. clearly sin 2x =0 or cos 2x=0 are not solutions. so we can divide by one of them.
3. (Original post by W.H.T)
Hence or otherwise show that the equation:

cosx (cos2x/cosx + sinx) = 1/2, can be written as:

sin2x = cos2x

afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
Divide both sides by cos2x to get tan2x =1

4. (Original post by W.H.T)
Hence or otherwise show that the equation:

cosx (cos2x/cosx + sinx) = 1/2, can be written as:

sin2x = cos2x

and afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
multiply out the brackets...
cos2x + sinx cosx = 1/2

then use this identity...
sin(2x) = 2 sinx cosx

and then to solve divide both sides by cos(2x), so sin(2x)/cos(2x) = tan(2x) = 1.
5. (Original post by kat_s)
multiply out the brackets...
cos2x + sinx cosx = 1/2

then use this identity...
sin(2x) = 2 sinx cosx

and then to solve divide both sides by cos(2x), so sin(2x)/cos(2x) = tan(2x) = 1.

for the first step of multiplying out the brackets, would I get:

cos^2 2x / cosx+sinx = 1/2
6. (Original post by Superman_Jr)
Divide both sides by cos2x to get tan2x =1

sin2x = cos2x

how can I divide by cos2x, because on the 'cos2x' side there is nothing multiplying it. So surely to get 'cos2x' to the other side, I would have to subtract by 'cos2x' ?
7. (Original post by W.H.T)
sin2x = cos2x

how can I divide by cos2x, because on the 'cos2x' side there is nothing multiplying it. So surely to get 'cos2x' to the other side, I would have to subtract by 'cos2x' ?
when you divide by cos2x, you'd be left with 1. And you can divide by cos2x as long as you do the same to both sides, the equation remains unchanged.
8. (Original post by W.H.T)
for the first step of multiplying out the brackets, would I get:

cos^2 2x / cosx+sinx = 1/2
cosx (cos2x/cosx + sinx) = 1/2

(cosx cos2x)/cosx + cosx sinx = 1/2

the cos x at the top and bottom of the first fraction cancel.

Unless you mean cosx (cos2x/(cosx +sinx)) for the original equation??
9. (Original post by kat_s)
cosx (cos2x/cosx + sinx) = 1/2

(cosx cos2x)/cosx + cosx sinx = 1/2

the cos x at the top and bottom of the first fraction cancel.

Unless you mean cosx (cos2x/(cosx +sinx)) for the original equation??

yeah my original equation is: cosx (cos2x/(cosx +sinx)) = 1/2

I don;t understand how you got the part in bold here:
(cosx cos2x)/cosx + cosx sinx = 1/2
10. how do you solve 2sin(2x)+4cos^2(x)=3

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