Find Coty

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

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

thanks.

Reply 1

Simba

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

(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.

Reply 2

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.

Reply 3

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.

No problem :smile:

...

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

Reply 4

generalebriety

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.

sinx / cosx = 2cosx / sinx
Multiply both sides by cosx and sinx.

Reply 5

Scholesey

ah,
had 2 tans and the penny didn't drop lol,
root 2 of course.

cheers.

Reply 6

Scholesey

hi,
sorry to be a pain, sitting here trying to do mixed exercise 6f in C3.

a. solve: sec(2@ - 15*) = cosec135* 0<@<360

and b. solve: cosec(x + pie/15) = -root2 0<x<2pie

thanks alot.

Reply 7

Simba

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(2@ - 15*) = cosec135*