If a third plane could be drawn which contains both lines a and b, then what must the lines be?
2 answers:
You might be interested in
The Lagrangian
has critical points where the first derivatives vanish:
We can't have , since that contradicts the last condition.
(0 critical points)
If two of them are zero, then the remaining variable has two possible values of . For example, if
, then
.
(6 critical points; 2 for each non-zero variable)
If only one of them is zero, then the squares of the remaining variables are equal and we would find (taking the negative root because
must be non-negative), and we can immediately find the critical points from there. For example, if
, then
. If both
are non-zero, then
, and
and for either choice of , we can independently choose from
.
(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)
If none of the variables are zero, then . We have
and similary have the same solutions whose signs can be picked independently of one another.
(8 critical points)
Now evaluate at each critical point; you should end up with a maximum value of
and a minimum value of
(both occurring at various critical points).
Here's a comprehensive list of all the critical points we found:
Answer:
x=πn, n∈Z
Step-by-step explanation:
if sin²2x=1-cos²2x, then
1-cos²2x-2cos2x+2=0; ⇒ cos²2x+2cos2x-3=0; ⇔ (cos2x+3)(cos2x-1)=0;