Question

factor and solve cot^2xcos^2x-cot^2x=0

Answer 1

\(\Large\color{black}{\cot^2x \cos^2x-\cot^2x=0}\)
So your first step, is to factor out of \(\Large\color{black}{\cot^2x }\), can you do that?

Answer 2

Here is another way to verify by working on both side together,
cot^2x - cos^2x = cot^2x * cos^2x, where cot^2x = (cos^2x)/(sin^2x)
Substitute cot^2x and multiply the second term with (sin^2x)/(sin^2x) for common denominator
(cos^2x)/(sin^2x) -( cos^2x)(sin^2x)/(sin^2x) = (cos^2x)/(sin^2x) * cos^2x
Multiply by sin^2x on both sides to eliminate denominator and factor, then
cos^2x(1 - sin^2x) = cos^2x * cos^2x
Devide by cos^2x on both sides
1- sin^2x = cos^2x, where 1 - sin^2x = cos^2x
Then, cos^2x = cos^2x, both sides same.

Answer 3

Yes, I factored it out

Answer 4

Okay, so I factored it to \[Cot ^{2}\alpha(\cos ^{2}\alpha)(-1)=0\]

Answer 5

\(\Large\color{black}{ \cot^2x \cos^2x - \cot^2x=0}\)

\(\Large\color{black}{ \cot^2x (\cos^2x - 1)=0}\)

Good? now apply the zero product property.

Answer 6

What's the zero product property?