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# A quick c3 trig question? watch

1. Find the solution of this equation:

Cot(x)= 3Sin2(x) interval -180<X<180

i got the worked solution, however its not making very much sense and I want to understand this.

Thanks
2. (Original post by Stanza)
3Sin2(x)
is that

or

?
3. (Original post by Pheylan)
is that

or

?
its the first one
4. its c2!

Heres an identity:
sin2(x)+cos2(x)=1

re-arrange:
sin2(x)=1-cos2(x)
sub in for sin2(x)

cos(x)=3(cos2(x))

then its easy
5. oh.

wow i read that wrong twice

cot (x)

also sin (2x)

sorry dude
6. (Original post by Stanza)
its the first one

7. (Original post by Jizkid Jnr)
its c2!

Heres an identity:
sin2(x)+cos2(x)=1

re-arrange:
sin2(x)=1-cos2(x)
sub in for sin2(x)

cos(x)=3(cos2(x))

then its easy
I think there is a misunderstanding. if I write it like this, maybe its better
cotØ=3sin2Ø and there is no squared, its a normal 2
8. (Original post by Stanza)
I think there is a misunderstanding. if I write it like this, maybe its better
cotØ=3sin2Ø and there is no squared, its a normal 2
9. yes thanks
10. (Original post by Stanza)
yes thanks
but the answer in the book says: square root of 1/6???????????????????
11. (Original post by Stanza)
but the answer in the book says: square root of 1/6???????????????????
Well sin(x) = square root of 1/6

3*(2sin(x)cos(x))= cos(x) / sin(x)

= 6 sin^2(x)cos(x) = cos(x)

= 6sin^2(x)=1 (dividing both sides by cos(x) )

=sin^2(x)= 1/6

sin(x) = root (1/6)
12. (Original post by Stanza)
Find the solution of this equation:

Cot(x)= 3Sin2(x) interval -180<X<180

i got the worked solution, however its not making very much sense and I want to understand this.

Thanks
cotx=3sin2x

1) write cotx as : cosx/sinx
2) And you know that sin2x is the same as 2sinxcos

therefore you have : cosx/sinx=3(2sinxcosx)
3) expand the rhs -- cosx/sinx=6sinxcosx
4)multiply both sides by sinx....this gives :

cosx=6sin^xcosx

5) now make it equal zero....

6sin^xcosx-cosx=0

6)But this can be facored since cosx is common to both
7)so.... cosx(6sin^x-1)=0

now its really easy.... cosx=0
0r 6sin^x-1=0

therefore arcos0=90

arcos(+/- .. (1/6))= 0

so you do arcos both plus and minus 1/6

remember that cos(360+/-x)=cosx
and sin(180-x)=sinx

is that correct>? or have i left something out?
13. (Original post by sabuuk)
is that correct>?
no
(Original post by sabuuk)
have i left something out?

14. (Original post by Pheylan)
no

lol...but he didn't want the answer, he just wanted some understanding

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