Q2: The Power of Triangles

First consider $n \geq 2$ .

Note that $\cos ^n x - \sin ^n x \leq |\cos ^n x - \sin ^n x |$ Now we apply the 'trick' that is the triangle inequality ; that is, for any reals $a,b$ , $|a+b| < |a| + |b|$ . Then: $|\cos ^n x - \sin ^n x | \leq |\cos ^n x| + |-\sin ^n x| = |\cos ^n x| + |\sin ^n x| \leq \cos ^2 x + \sin ^2 x = 1$ Therefore $\sin^2 x = |\sin ^n x|, \cos^2 x = |\cos ^n x|$ .

From this we find the solutions $0,\pi,\frac{3\pi}{2}$ .

In the case of $n = 1$ , we have $\cos x - \sin x = 1$ , and applying the R method we find that $\sin(x - \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$ , giving the solutions $0, \frac{3\pi}{2}$ .

Therefore there are $\boxed{3}$ unique solutions in total.

$Cos^nX-Sin^nX=1.\\ \therefore~if~Cos^nX,Sin^nX ~has~irrational~component,~Cos^nX,Sin^nX~must~be~of~the~type,~a\pm\sqrt[p]m ~and~b\pm\sqrt[p]m,~~a, b\neq 0.\\ \text{There is no X that satisfies this.......(X=45, a=b=0).}\\ Let~ Cos^nX,~ and~ Sin^nX \text{ be pure fractions. They must have a common denominator and still } Cos^nX- Sin^nX=1.\\ There~is ~no~such~X.\\ So~ the~ only~ possibility~ is ~~~Cos^nX=0 ~~or~ ~1~~~~~ with~~~~~ Sin^nX= - 1~~ or~~ 0.\\ Cos^nX=0~~and~~Sin^nX= - 1, ~when~X=270 ~~~for~ all~ odd~ n.\\ Cos^nX=1~~and~~Sin^nX= 0, ~when~X=0~~OR~~X=180~~for~all~even~ n.\\$
Three solutions for X=0, X=180, X= 270.