Evaluate x, given:
(sin x)^8+(cos x)^8 = sqrt (2- sin x)
using pen and paper only,
Consider the LHS and RHS as separate functions, f(x) and g(x).
Let s = sin x and c = cos x
LHS: f(x) = s^8 + c^8
derivative is f'(x) = 8*s*c(s^6 - c^6) = 0 if:
s = c which occurs at x = pi/4 + n*pi (these are the minima)
or s = 0 or c = 0 which occurs at n*pi/2 (these are the maxima)
f(pi/4) = 2*(sqrt(2)/2)^8 = 1/8 which is a minimum for f.
Clearly if sin or cos = 1, the other is 0, and fmax = 1, when x = n*pi/2
1/8 <= f(x) <= 1
RHS: g(x) = sqrt(2 - sin x) has a min and max when sin x = 1 and -1
1 <= g(x) <= sqrt(3)
g(x) = 1 when sin(x) = 1, or x = pi/2 + 2n*pi
The only places RHS = LHS is at max f(x) and min g(x)
Both the max for f and the min for g occur at:
x = pi/2 + 2n*pi
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Posted by Larry
on 2023-03-06 13:47:02 |
Solution
