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Do you see the equation editor in the bottom left? that would be good to use

\[\sqrt{x+3}/(x+8)(x-2)\]

f(x)= ^

no not sure.

if you add -3 to both sides then you get x>-3?

oh I see it, wow your good.

(2+8)(2-2)

2-2=0

0 times anything = 0

can we do another?

ok

Verify the identity.
cos 4x + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x

It looks painful. Do you have the "half-angle" formula for cos 4x ?

isn't it =+-square root 1+ cos theta/ 2

\[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = ? \]

2-2sin^2(a) ?

\[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = 1 - 2 \sin^2(2x) \]

so 1-2sin^2(2x)
1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x

yes, and if we add +2sin^2(2x) to both sides, we can simplify that a bit

yes

oops, you should add 2 sin^2(2x) (not 2sin^2(x) )

It looks like we should use the same rule as before , but now on cos(2x)

1-2sin^2(x)

yes, put that in for cos(2x)

the left side is
\[ 1 + 1 - 2\sin^2(2x) \]

yes, both sides are obviously the same.

can we do another problem, not like this one though.

one more, but please make it a new post.

ok hold on.