What is the solution for #cos^nx-sin^nx=1# witn #n in NN^+#?
2 Answers
Explanation:
Let me solve
Squaring both sides,
So,
#(sinx cos x)^n=0 to( sin (2x))^n=0 to sin (2x)=0 to 2x=kpi to x = k/2pi,
k=0, +-1, +-2, +-3, ...#. This includes extraneous solutions for odd k
against even n.#
With this clue, let us consider the given problem.
Rearranging as
Rearranging,
So,
Here, odd k gives extraneous solution for odd n.
To include these cases as well, let us try squaring the alternative
rearrangement.
k=+-1, +-3, +-5.... n=1, 3, 5, ..#
If
If
Explanation:
For
We know that
Now supposing
but
now supposing
but
so the equality is observed only when
and the solutions are
Ressuming
If
If