Algebra trig question

Please solve (and show as muck work as you can/ explain your reasoning)-

1. Find all values for which the impulses y=2sin^2 x and y=1-sin x meet in the interval 0 deg. is less than or equal to 360 deg.

2.solve for x: log3(x^2-4)-log3(x+2)=2

Much appreciated!

1.
y=2sin² x
and
y= 1 - sin x

so

2sin² x = 1 - sin x

thus

2sin² x + sin x - 1 = 0

let u = sin x then

2u² + u - 1 = 0
implies
(2u -1)(u+1) = 0

so 2u = 1 →u= ½
or

u= -1
but u = sin x

so sin x = ½ implies that x= 30° or x = 150°
and
sin x = -1 implies that x= 270°

so solutions between 0° and 360° are {30°,150°,270°}


2. log (x² - 4) - log(x+2) = 2
.........3................3

implies
(x+2)(x-2)
log▬▬▬▬▬ = 2
....3....(x-2)

now we must note x is not equal to -2 since that would put a zero in denominator
and an unallowed zero in argument of log function

we get

log (x+2) = 2
....3

so 3² = x+2 writing this in exponential form

and this simplifies to 9 = x+2

subtract 2 from both sides to get
x=7

question 1:
if you make y1 and y2 equal each other you will get a quadratic equation such as: (2sinx-1)(sinx+1)=0
when you solve that you get (II/6, 5II/6,7II/6,11II/6, and 3II/2).
question2 is pretty simple. read the section on your book you will get it easily.