Need help with finding critical numbers of extrema

[rayroshi]

Okay, I am new to this site, and am a total beginner, trying to learn calculus on my own. I am using "Calculus for Dummies," which seems like a good book. For those of you who have the book, I am trying to figure out how the author got the answers that he did, in the middle of page 164, where he found the critical numbers of the function. I understand everything up to, and including, the very last part of step 1, where he determines that x= 3pi/2, 7pi/6, and 11pi/6. I just don't see where these numbers came from.

For those of you who don't have the book, here is a step-by-step breakdown, just as he shows it in the book:
If h(x) = cos(2x) - 2sinx, and we are to find the critical numbers of the function within the closed interval [pi/2, 2pi], then
h'(x) = -sin(2x) * 2 - 2cosx (chain rule)
0 = -2sin(2x) - 2cosx
0 = sin(2x) + cosx (divide both sides by -2)
0 = 2sinxcosx + cosx (trig identity)
0 = cosx(2sinx + 1) (factor out cosx)
so, cosx = 0 or 2sinx + 1 = 0
If cosx = 0, then x=3pi/2 (and here's my first question: where'd the "3pi/2" come from?)
If 2sinx + 1 = 0, then x = 7pi/6, 11pi/6 (and that's my second question: where'd 7pi/6 and11pi/6 come from?)

Any help would be much appreciated!

 

[Deleted member 4993]

Okay, I am new to this site, and am a total beginner, trying to learn calculus on my own. I am using "Calculus for Dummies," which seems like a good book. For those of you who have the book, I am trying to figure out how the author got the answers that he did, in the middle of page 164, where he found the critical numbers of the function. I understand everything up to, and including, the very last part of step 1, where he determines that x= 3pi/2, 7pi/6, and 11pi/6. I just don't see where these numbers came from.

For those of you who don't have the book, here is a step-by-step breakdown, just as he shows it in the book:
If h(x) = cos(2x) - 2sinx, and we are to find the critical numbers of the function within the closed interval [pi/2, 2pi], then
h'(x) = -sin(2x) * 2 - 2cosx (chain rule)
0 = -2sin(2x) - 2cosx
0 = sin(2x) + cosx (divide both sides by -2)
0 = 2sinxcosx + cosx (trig identity)
0 = cosx(2sinx + 1) (factor out cosx)
so, cosx = 0 or 2sinx + 1 = 0
If cosx = 0, then x=3pi/2 (and here's my first question: where'd the "3pi/2" come from?)
If 2sinx + 1 = 0, then x = 7pi/6, 11pi/6 (and that's my second question: where'd 7pi/6 and11pi/6 come from?)

Any help would be much appreciated!

cos(x) = 0

cosine function has roots (is equal zero) at x = pi/2 and 3pi/2 and 5pi/2 and so on.

Within the given domain [pi/2, 2pi], cos(x) = 0 is at x = 3 pi/2

similarly

2*sin(x) + 1 = 0

sin(x) = - 1/2

and sin(x) will be -1/2 for those given values.

 

[mmm4444bot]

where'd the "3pi/2" come from?

Three halves of Pi is one of the many special radian angles that we memorize in trigonometry.

Trigonometry is a prerequisite topic for students of calculus IF their calculus covers trigonometric functions.

Google keywords special angles for trig