would it be...
\frac{\sin^2\theta+\cos^2\theta}{\cos\theta\sin\theta}
then
\frac{1-\cos^2+\cos^2\theta}{\cos\theta\sin\theta}
into
\frac{1}{\cos\theta\sin\theta}
i feel so frustrated :confused: sry
im sorry I am really bad with these i just started doing them um so i do
\frac{\sin^2\theta+\cos\theta}{\cos\theta\sin\theta}
i still don't get it won't that make it more complicated? :confused:
so i get...
\sin\theta\frac{1}{\cos\theta}+cos\theta\frac{1}{\sin\theta}
then i put each in one fraction right?
\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}
to
\tan\theta+cot\theta
is that the simplest it can get?
cos(x)-\frac{cos^{2}(x)}{cos(x)-sin(x)}
after long staring and thinking a light bulb just lit in my head lol
so...
\frac{\cos^2(x)-\sin(x)\cos(x)-\cos^2(x)}{cos(x)-sin(x)}
to
\frac{-\sin(x)\cos(x)}{cos(x)-sin(x)} then all multiplied by -1 equals...
oops i put in the problem wrong and i figured it out(the second one)
f(x)=1/2x^3-x
x(1/2x^2-1)=0
x=0
1/2x^2-1=0
1/2x^2=1
x^2=2
x= +-square root of 2
Thank you for your assistance.
Homework Statement
Find the zeros of the function algebraically
Homework Equations
f(x)=x/(9x^2-4)
and
f(x)=x^3-x
The Attempt at a Solution
i was gone for about a month and a half from the class because i had to move and I am kinda rusty
ATTEMPT-
f(x)=x/(9x^2-4)
Multiplied by...