Question

i have two questions for you to check please!

6cos^2x+5cosx-4=0 where 0degrees < or equal to 0 which is < or equal to 360.
I got that this factors to (3x+4)(2x-1) so the answer is x= -4/3 and x= 1/2 ?? right?

and the second question:

determine sin2theta, cos2theta, and tan2theta. the given is sin x=3/5 and that cosx < 0. (this is where i get confused. how do i know what quadrant its in if cosx<0?)

If i put it in first i get 5 as hypotnuse 4 as adjacent side and 3 as opp? is that going in the right direction? please help

for the first you are not done
You are on the right track to treat it as a quadratic.
your factors are (3cosx +4)(2cosx - 1)=0
so cosx =-4/3 or cosx = 1/2
the first of these is not possible since the cosine lies between -1 and +1.
so cosx = 1/2
the cosine is positive in quadrants 1 and 4
so if cosx=1/2 <b>x = 60º or x = 300º</b>

for you second question....
<<given is sin x=3/5 and that cosx < 0>>

If you know the CAST rule, then the angle must be in quadrant #2 (between 90 and 180)

if cosx=3/5, you were right to draw a triangle and conclude that sinx must be 4/5

Now use the special angle relationships
sin 2x = 2sinxcosx = 2(4/5)(3/5) = ...
cos 2x = cos^2 x - sin^2 x = 9/25 - 16/25 = ...

and of course tan 2x = sin 2x/cos 2x so....

(pi/3 and 5pi/3 in radians)

That should have been

sin x = 3/5 and cosx = -4/5

I am sure you can make the necessary changes in my substitution

Also the "(pi/3 and 5pi/3 in radians)"
should have been at the end of the first solution.