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# C3 question watch

1. Can anybody help me solve this equation:

sin2x - tanx = 2cos2x
for x in the interval 0<= x <= 180 degrees. (Greater than or equal to 0 and less than or equal to 180 degrees, if unclear).

So far, I have:
4cosx - 4cos^3 + 2sinxcos^2 -sinx -2cosx = 0

2. (Original post by UserXi1.fr)
Can anybody help me solve this equation:

sin2x - tanx = 2cos2x
for x in the interval 0<= x <= 180 degrees. (Greater than or equal to 0 and less than or equal to 180 degrees, if unclear).

So far, I have:
4cosx - 4cos^3 + 2sinxcos^2 -sinx -2cosx = 0

Try this

2sincos^2 - sin = 2cos(2cos^2-1)

Which is a step before where you were

Factorise the LHS
3. These were my individual steps:
1)sin2x - tanx = 2cos2x
2)2sinxcosx - tanx = 2(1- 2sin^2x)
3) 2sinxcosx -tanx -2 +4sin^2x = 0
4) 4sin^2 + 2sinxcosx - tanx -2 = 0
5) 4sin^2 + 2sinxcosx - sinx/cosx -2 = 0
6) 4sin^2 cosx + 2sinxcos^2 - sinx - 2cosx = 0
7) 4(1-cos^2)cosx +2sinxcosx -sinx -2cosx =0
8) 4cosx -4cos^3 +2sinxcos^2 -sinx -2cosx= 0

That's as far as I got, but with a variety of sinx and cosx, any obvious errors?
4. I got 2sin^2(x)-sin(x)-1=0

is this correct?
5. tan x cos 2x = 2 cos 2x
cos 2x (tan x − 2) = 0 M1
cos 2x = 0 or tan x = 2 A1
2x = 90, 270 or x = 63.4 B1
x = 45°, 63.4° (1dp), 135°

That is the official answer in the mark scheme
6. (Original post by UserXi1.fr)
These were my individual steps:
1)sin2x - tanx = 2cos2x
2)2sinxcosx - tanx = 2(1- 2sin^2x)
3) 2sinxcosx -tanx -2 +4sin^2x = 0
4) 4sin^2 + 2sinxcosx - tanx -2 = 0
5) 4sin^2 + 2sinxcosx - sinx/cosx -2 = 0
6) 4sin^2 cosx + 2sinxcos^2 - sinx - 2cosx = 0
7) 4(1-cos^2)cosx +2sinxcosx -sinx -2cosx =0
8) 4cosx -4cos^3 +2sinxcos^2 -sinx -2cosx= 0

That's as far as I got, but with a variety of sinx and cosx, any obvious errors?
If you use 2cos^2 - 1 instead in line 2 you have what I had which factorised as the mark scheme shows
7. is this edxcel,if so which paper
8. http://www.freeexampapers.com/past_p...FC3%2FSolomon/

Paper E, Q 2B
9. (Original post by TenOfThem)
If you use 2cos^2 - 1 instead in line 2 you have what I had which factorised as the mark scheme shows
I've used that... yet to no avail. Any complete solutions for the answer, mark scheme seems to skip a few steps.
10. (Original post by UserXi1.fr)
I've used that... yet to no avail. Any complete solutions for the answer, mark scheme seems to skip a few steps.
Do you have what I put in post 2

Can you factorise the LHS
11. 2)2)2sinxcosx - tanx = 2(1- 2sin^2x)

Yeah so if I replaced (1-2sin^2 x) with (2cos^2 x -1), so

2sinxcosx - sinx/cosx = 2( 2cos^2 x -1)
2sinxcos^2 - sinx = 2(2cos^2 -1).

Stuck here, tried it multiple times, end up expanding and getting cosx and sinx all over the place!
12. Got 2/3 solutions.

1) 2sincos^2 -sinx= 2(2cos^2 -1)
2) sinx(2cos^x-1) = 2(2cos^2-1)
3) sinxcos2x -2cos2x = 0
4) cos2x( sinx-2)=0
5) cos2x=0 {sinx=2... no solutions}
6) 2x= 90, 270.....x=45, 135.
Missing a 63.4 degrees though.
13. (Original post by UserXi1.fr)
Got 2/3 solutions.

1) 2sincos^2 -sinx= 2(2cos^2 -1)
2) sinx(2cos^x-1) = 2(2cos^2-1)
3) sinxcos2x -2cos2x = 0
4) cos2x( sinx-2)=0
5) cos2x=0 {sinx=2... no solutions}
6) 2x= 90, 270.....x=45, 135.
Missing a 63.4 degrees though.
You are missing a cos in line 2
14. (Original post by UserXi1.fr)
Can anybody help me solve this equation:

sin2x - tanx = 2cos2x
for x in the interval 0<= x <= 180 degrees. (Greater than or equal to 0 and less than or equal to 180 degrees, if unclear).
I think the easiest method is to form a cubic in tan x.

Divide each term by

Use the Pythagorean identity and you should be home.
15. (Original post by Mr M)
I think the easiest method is to form a cubic in tan x.

Divide each term by

Use the Pythagorean identity and you should be home.

Good, found the solutions using both methods.
Thanks all!

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