Trig solutions question (1 Viewer)

[tsoliman1]

(Sinx-1)(tanx+2)=0

How many solutions are there between 0 and 2pi ?

I got 3 solutions but the answers says 2.

 

[InteGrand]

(Sinx-1)(tanx+2)=0

How many solutions are there between 0 and 2pi ?

I got 3 solutions but the answers says 2.

Note x = pi/2 is void because the tan(x) in the LHS is undefined at pi/2 (so pi/2 isn't part of the domain of the LHS).

(You may find it interesting to investigate what happens to the LHS as x approaches pi/2.)

 

[tsoliman1]

Omg How did I not realise that. I knew it was something like that.

Thx a lot

 

[tsoliman1]

Also if x approached Pi/2 wouldn't there be a vertical asymptote at x=Pi/2

 

[InteGrand]

Also if x approached Pi/2 wouldn't there be a vertical asymptote at x=Pi/2

It turns out no. The limit is actually 0. An easy way to show this is with the help of L'Hôpital's rule (which isn't in the syllabus though).

Here's the graph of the function:

http://wolframalpha.com/input/?i=plot+y+=+(sin+x+-1)(tan+x++2)&x=0&y=0 .

Since it approaches 0, we could extend the function to a continuous one defined on the reals by defining it to be 0 at these places where tan(x) blows up. If we did this, the answer to the question would have been three instead (if the Q. was rephrased as solving f^*(x) = 0 in the same interval 0 to 2pi, where f^* is the described continuous extension of (sin(x) – 1)(tan(x) + 2)).