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method for :

5sinxcosy + 4cosxsiny = 0 , cotx= 2 ; find cot y

i know cot = cos/sin, but unsure when sincos' are together.

thanks.

5sinxcosy + 4cosxsiny = 0 , cotx= 2 ; find cot y

i know cot = cos/sin, but unsure when sincos' are together.

thanks.

Scholesey

method for :

5sinxcosy + 4cosxsiny = 0 , cotx= 2 ; find cot y

i know cot = cos/sin, but unsure when sincos' are together.

thanks.

5sinxcosy + 4cosxsiny = 0 , cotx= 2 ; find cot y

i know cot = cos/sin, but unsure when sincos' are together.

thanks.

5sinxcosy + 4cosxsiny = 0

(sin x)(5cos y + 4cot x.sin y) = 0

(sin x)(sin y)(5cot y + 4cot x) = 0

(sin x)(sin y)(5cot y + 8) = 0

So, either:

sin x = 0, sin y = 0 or 5cot y + 8 = 0.

Looking at the last one:

5cot y + 8 = 0

cot y = -1.6.

Hope this helps,

~~Simba

Edit: Actually, because we know cot x = 2, some of our solutions can be discarded, but we don't need to worry about that for this problem I don't think.

Scholesey

thanks alot,

and just one quick one,

i cant seem to get the right angles, even though it looks simple, i must be making it too complicated :

solve tanx=2cotx -180<x<90

thanks again.

and just one quick one,

i cant seem to get the right angles, even though it looks simple, i must be making it too complicated :

solve tanx=2cotx -180<x<90

thanks again.

No problem ...

tan x = 2/tan x

tan²x = 2

tan x = root 2

x = 54.7356...°, -125.2643...°

= 54.7°, -125.3° (1 d.p.)

Hope this helps,

~~Simba

Scholesey

thanks alot,

and just one quick one,

i cant seem to get the right angles, even though it looks simple, i must be making it too complicated :

solve tanx=2cotx -180<x<90

thanks again.

and just one quick one,

i cant seem to get the right angles, even though it looks simple, i must be making it too complicated :

solve tanx=2cotx -180<x<90

thanks again.

sinx / cosx = 2cosx / sinx

Multiply both sides by cosx and sinx.

Perhaps you should ask your teacher to go over a couple of questions on the board as examples if you're having trouble with a lot of them. It's probably more benificial than just posting them all on here.

sec(2x - 15°) = root 2.

1/cos(2x - 15°) = root 2

cos(2x - 15°) = (root 2)/2

2x - 15° = 45°, 315°, 405°, 675°.

=> x° = 30°, 165°, 210°, 345°.

sin(x + pi/15) = -(root 2)/2

x + pi/15 = 3pi/4, 5pi/4.

=> x = 41pi/60, 71pi/60.

Hope these help,

~~Simba

sec(2@ - 15*) = cosec135*

sec(2x - 15°) = root 2.

1/cos(2x - 15°) = root 2

cos(2x - 15°) = (root 2)/2

2x - 15° = 45°, 315°, 405°, 675°.

=> x° = 30°, 165°, 210°, 345°.

cosec(x + pie/15) = -root2

sin(x + pi/15) = -(root 2)/2

x + pi/15 = 3pi/4, 5pi/4.

=> x = 41pi/60, 71pi/60.

Hope these help,

~~Simba

- Accommodation - finding potential flatmates?
- Muslim student accommodation in Cambridge
- UCAS id
- Do you find The Student Room intuitive to use?
- A quick clarification on vectors, acceleration and forces
- Where do Queen Mary students live in second and third year?
- equations of tangents and normals
- Unique learner number help ?
- LGBT at liverpool uni
- What to do with myself
- Friends!!
- Media Studies (9607) for Cambridge International A-Levels, which textbook do you use?
- Buddhists in Chester?
- flatmates
- Where can I get A and AS level exams?
- Where to find exam questions on NCFE health and social care level 1/2
- Finding my Heritage
- Marriage
- Nottingham Trent University Clearing 2024
- Accommodation at Broadgate Park (near University Park)

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