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Solve for x in -sin(x)8cos(x)=0?

GROUPS:
Not that good with trig expressions and Im having trouble looking at these in an intuitive sense. This is actually from a problem that asks for the critical points, absolute extreme values and intervals on which f is increasing and decreasing for the function f(x)=4cos^2(x) on the interval [0,pi].

Okay, so since this sam person is, in my opinion, acting pretty condescending... I will do the same. You see, for this particular problem the given answers from Wolfram Alpha are pretty useless to me... at least in my interpretation of them. I know this because I have already, in fact, put them into Wolfram Alpha, and even my copy of Mathmatica. Why you would think I came to the wolfram forum and asked a question before I copied and pasted a problem into WA is beyond me. Second of all, thank you very much to everyone ELSE that answered... But the actual answer turned out to be to set each part to 0 since each much equal 0 at some point. After doing that you get pi/2 as the only point on that interval that isnt an end point. 
POSTED BY: Andrue Stafford
Answer
9 months ago
Sometimes plotting the curves helps.

Red is the original; Blue is the derivative.


You might check the documentation for terms like "solve" or "equation solving".
POSTED BY: Bruce Miller
Answer
9 months ago
Reduce[Sin[x] Cos[x] == 0, x]
C[1] \[Element]
  Integers && (x == -(\[Pi]/2) + 2 \[Pi] C[1] ||
   x == \[Pi]/2 + 2 \[Pi] C[1] || x == 2 \[Pi] C[1] ||
   x == \[Pi] + 2 \[Pi] C[1])
POSTED BY: Frank Kampas
Answer
9 months ago
Andrue, I just literally copied and pasted your *title* into Wolfram|Alpha and got the following answer - see image below or click on ithe linked text. This took me less time than for you to type your post. Now the real question is - is it homework? You should always show in your post *what have YOU tried* - using Mathemaitca or Wolfram|Alpha or other Wolfram tools.

POSTED BY: Sam Carrettie
Answer
9 months ago
Reduce[-8 Sin[x] Cos[x] >= 0, x, Reals]
C[1] \[Element]
  Integers && (x == \[Pi] + 2 \[Pi] C[1] ||
   1/2 (-\[Pi] + 4 \[Pi] C[1]) <= x <= 2 \[Pi] C[1] ||
   1/2 (\[Pi] + 4 \[Pi] C[1]) <= x < \[Pi] + 2 \[Pi] C[1])
Reduce[-8 Sin[x] Cos[x] <=  0, x, Reals]
C[1] \[Element]
  Integers && (x == \[Pi] + 2 \[Pi] C[1] || -\[Pi] + 2 \[Pi] C[1] <
    x <= 1/2 (-\[Pi] + 4 \[Pi] C[1]) ||
   2 \[Pi] C[1] <= x <= 1/2 (\[Pi] + 4 \[Pi] C[1]))
POSTED BY: Frank Kampas
Answer
9 months ago