In the range \(0\le x < 2\pi\) the equation \[\sin^8 x + \cos^6 x = 1\] has
\(3\) solutions,
\(4\) solutions,
\(6\) solutions,
\(8\) solutions.
There are at least two ways we could tackle this problem:
Method 1
It’s easy to see that
- \(\sin x=0, \cos x=1\) at \(x = 0\),
- \(\sin x = 0, \cos x = -1\) at \(x = \pi\),
- \(\sin x = 1, \cos x =0\) at \(x = \pi/2,\) and
- \(\sin x = -1, \cos x = 0\) at \(x = 3\pi/2\),
so we definitely have four solutions.
Are there any solutions where \(\sin x\) and \(\cos x\) are both non-zero?
In this case, we know that \(0 < \sin^2x < 1\) and \(0 < \cos^2x < 1\).
Now for any \(y\) between \(0\) and \(1\), \(y^4<y\) and \(y^3<y\).
So we know that \[\sin^8x=(\sin^2x)^4<\sin^2x, \cos^6x=(\cos^2x)^3<\cos^2x.\]
Thus \[\sin^8x+\cos^6x<\sin^2x+\cos^2x=1.\]
So our equation only holds for our four cases above, and there are exactly four solutions; the answer is (b).
Method 2
For convenience, let’s write \(s=\sin x\) and \(c=\cos x\).
\[\begin{align*} &s^8+c^6=1\\ \iff\quad& s^8+(1-s^2)^3=1\\ \iff\quad& s^8-s^6+3s^4-3s^2+1=1\\ \iff\quad& s^2(s^6-s^4+3s^2-3)=0\\ \iff\quad& s^2(s^2-1)(s^4+3)=0. \end{align*}\](We could equivalently substitute \(s^2 = 1 - c^2\).)
So there are solutions when \(s^2=0\), \(s^2=1\) and \(s^4=-3\) (the last has no real solutions).
So we require \(\sin x=0\) or \(\sin x=\pm 1\).
As in the method above, the solutions in the relevant range are \(x=0, \pi, \pi/2, 3\pi/2\), and the answer is (b).
