Question

solve (sin^2 x)(cos3x)+(cos3x)(cos^2 x)=1/2 for x over [0,2pi)

Answer 1

factor cos 3x out. you can see the solution.

Answer 2

ok but how do i do the inverse thing?

Answer 3

cos = 1/2 at pi/3 right? so would that be the answer?

Answer 4

cos (3x) = 1/2 . For convenient, let 3x = t , so cos t = 1/2 in the interval [0,2pi]
you have t = pi/3 and t = 5pi/3
now replace t = 3x so,
\[3x = \frac{\pi}{3}\rightarrow x=\frac{\pi}{9}\]do the same with 5pi/3

Answer 5

oh ok thank you so much!11

Answer 6

got me? good

Answer 7

\[y=\frac{ \sqrt{x ^{4}+6x ^{2}+9}}{2x ^{2}-10 }\]

Answer 8

solve it?

Answer 9

do you know how to find all of the asymptotes?

Answer 10

do you know what the asymptotes mean?

Answer 11

i know what it looks like in a graph, i just don't know how to solve and find them

Answer 12

x=0 and y=0

Answer 13

good, with x =0, the function undefined, right? with y =0, the same.
so, with the fraction as yours, when it's undefined? is not that when the denominator =0?

Answer 14

oh ok, so if i set the denominator = 0 then i can find the asymptotes

Answer 15

now, go backward, when denominator =0, the fraction is undefined and the value make it undefined is its asymptotes. YOU GOT IT

Answer 16

wud wud, happy when you got it quickly.

Answer 17

ha!! you see , my sentence sounds like a poem, right? hahahaha. I am day dreaming.

Answer 18

haha yay! ok is that all there is? do i have to do anything different if there is a removable discontinuity or is it any different because there is a square root in the numerator?

Answer 19

just consider the denominator if the question is about asymptote. the numerator is never negative for all x

Answer 20

ok thanks!!!

Answer 21

yw