I need some help with the domains of trig equations.
I have solved all the equations i believe with the right answers, , you can recheck it if you feel comfortable.

a. cos2x+cosx=0
ANS:
2cosx^2+cosx-1
(2cosx-1)(cosx+1)
x = 1/2 and x = -1

x = pi/3 ,5pi/3 and -pi

are these the right answers in the domain x is an element [-pi,pi) - i believe the [ means equal to and the bracket means doesnt equal

b. 2cosx^2+3sinx=0
2(1-sinx^2)+3sinx=0
2-2sinx^2+3sinx
2sinx^2-3sinx-2
(2sinx-1)(sinx+2)
sinx=1/2 sinx=-2 - cancels since -2 isnt on the unit circle
x = 30 and 150 degrees -- in the domain [0,360]

c. cos2x-cosx^2=0
2cosx^2-1-cosx^2
cosx^2-1
(cosx+1)(cosx-1)
cosx=-1 cosx= 1
x = pi and 0 ---- in the domain [-pi,pi]

i just have a hard time with these domains for the trig equations, can someone please check over my answers to see if it fits the domains, or if i have to get rid of some answers or add some answers.

proving identities.

cosx-sinx/cosx = 1-tanx

can someone prove this identity for me?

1.)

a) 5pi/3 is not in the given interval. You need to find the equivalent angle that is.

b) You factored the quadratic in sin(x) incorrectly.

c) You are missing a solution.

Identity: assuming you meant to type (cos(x) - sin(x))/cos(x) = 1 - tan(x)

Use the property of fractions $\frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}$ then use $\tan(x)\equiv\frac{\sin(x)}{\cos(x)}$.

well thank you for pointing out my errors but can you be more specific?

i did them to the best of my ability and thank you for the proving identites

i figured out my error in b its -2 and +1
can you please help me with the domain, telling me its just wrong doesnt help me much, sorry if that sounds rude but im having some major difficulty with the - to + domains

for a) would it be just pi/3 and pi?
and for b is it 300 degrees an 210?

For a), 5pi/3 is a good angle, you just need to subtract 2pi from it to get an equivalent angle in the given interval. Or use cos(-x) = cos(x).

For b) it would be 210° and 330°. See you want (180+30)° and (360 - 30)°.

oops i made the mistake of thinkin i was a tall triangle for a second

so it would become -pi/3, pi/3 and negative pi? since i cannot use pi since pi is not included in the domain, or would negtiave pi be negated as well.

You can use -pi, but not pi since is is not part of the interval.

thank you very much, if you dont mind me asking one more question.

If <a and <b are both in quadrant 3, sina=-3/5 cosb=-5/13

find tan(a-b)

i did tan(5-13)
(tan5-tan13)/(1+(tan5*tan13)) = (5-13)/(1+(5*13)) = -8/66 = -4/33

is this correct? it seems off

I would use the fact that sine and cosine are both negative in quadrant 3 to state:

$\cos(a)=-\sqrt{1-\sin^2(a)}=-\frac{4}{5}$

$\sin(b)=-\sqrt{1-\cos^2(b)}=-\frac{12}{13}$

and so:

$\tan(a-b)=\frac{\sin(a-b)}{\cos(a-b)}=\frac{\sin(a)\cos(b)-\cos(a)\sin(b)}{\cos(a)\cos(b)+\sin(a)\sin(b)}=$

$\frac{- \frac{33}{65}}{ \frac{56}{65}}=- \frac{33}{56}$

hm thats interesting never apporached it like that how come i get a different answer using the formula

tan(a-b) = tanA-tanB / 1+tanAtanB

did i substitute the numbers wrong? tan is the hyp if im correct.

I was under the impression that any and all of these forumlas that are related are supposed to give you the same answer