Expert: Paul Klarreich - 12/15/2011

Question
Hi, I'm hoping you'll be able to help me out with some problems...
The questions states:  solve for all values of the variable over the specified range:
a)  (sin2x)(cos2x+1) = 0  (0<x<360)
This is what I have so far:
let 2x = y, therefore siny=0 & cosy=-1
Then I can solve for when siny=0 by the sine graph, so it equals 0 at 0, 180 & 360 degrees.  But y = 2x, so x = 0, 90 & 180 degrees.  I can go through the same process w/ cos & I get x = 315.  Okay, so I have 4 answers.  But now I know that there are two other answers:  x=270 & 360.  How do I get these & how do I know how to look for these?

Answer
Questioner: Mike
Country: Ontario, Canada
Category: Calculus
Private: No
Subject: TRIG Solutions
Question: Hi, I'm hoping you'll be able to help me out with some problems...
The questions states:  solve for all values of the variable over the specified range:
a)  (sin2x)(cos2x+1) = 0  (0<x<360)
This is what I have so far:
let 2x = y, therefore siny=0 & cosy=-1
Then I can solve for when siny=0 by the sine graph, so it equals 0 at 0, 180 & 360 degrees.  But y = 2x, so x = 0, 90 & 180 degrees.  I can go through the same process w/ cos & I get x = 315.  Okay, so I have 4 answers.  But now I know that there are two other answers:  x=270 & 360.  How do I get these & how do I know how to look for these?
......................................
You did Ok,so far.  But I think that when you start with:

let  y = 2x.

YOU SHOULD SAY:

If      0 <= x <= 2pi,
then    0 <= 2x <= 4pi,
and so: 0 <= y <= 4pi

NOW when you solve for y by looking at the graph, you must look at  y = 0 to 4pi;  you will have two cycles and will see solutions to sin y = 0 at y = 0, pi, 2pi, 3pi, 4pi.

Same thing for  cos y = -1.

y = pi, 3pi.

This actually duplicates the others.  I think your  x = 315 is wrong.

BTW:  Get accustomed to working in radian measure.  Degrees are for children.