Question

Solve the equation algebraically: tanxsinxcosx - 1 = 0

Now, can someone explain why the answer has no solution? Haalllppppp

Answer 1

perhaps if we rewrite
\[\tan(x)\cos(x)\sin(x)\] it will be more clear

Answer 2

since
\[\tan(x)=\frac{\sin(x)}{\cos(x)}\] you are looking at
\[\frac{\sin(x)}{\cos(x)}\sin(x)\cos(x)=\sin^2(x)\]

Answer 3

solving
\[\sin^2(x)-1=0\] is not hard
what do you get?

Answer 4

So we don't factor it like
\[(\sin(x)-1) (\sin(x)+1) = 0?\]

Answer 5

you can if you like

Answer 6

or you can just write
\[\sin^2(x)=1\]so
\[\sin(x)=1\] or
\[\sin(x)=-1\]

Answer 7

idk apparently the answer doesn't have a solution, that's where i got confused :/

Answer 8

we are not done yet!

Answer 9

LOL

Answer 10

once we solve
\[\sin(x)=1\] for \(x\) we can see why there is no solution
did you solve it yet?

Answer 11

How so? I thought the solution can also be \[\pi/2\] or \[3\pi/2\]

Answer 12

so that makes it undefined?

Answer 13

right

Answer 14

since at those number, cosine is zero

Answer 15

don't forget you stared with
\[\tan(x)\sin(x)\cos(x)\] and since cosine is zero at those number,
a) tangent is not defined there and
b) cosine is zero, so you would actually have \(\frac{1}{0}\times 1\times 0\)
moral of the story
you cannot cancel zeros

Answer 16

ahah, that makes a lot of sense now hehe :) tysm!