When is $\sin^8 x+\cos^6 x$ equal to $1$?

When is $\sin^8 x+\cos^6 x$ equal to $1$?

Question

In the range $0\le x < 2\pi$ the equation $\sin^8 x + \cos^6 x = 1$ has

1. $3$ solutions,

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1. $4$ solutions,

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1. $6$ solutions,

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1. $8$ solutions.

Solution

There are at least two ways we could tackle this problem:

Method 1

It’s easy to see that

1. $\sin x=0, \cos x=1$ at $x = 0$,

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1. $\sin x = 0, \cos x = -1$ at $x = \pi$,

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1. $\sin x = 1, \cos x =0$ at $x = \pi/2,$ and

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1. $\sin x = -1, \cos x = 0$ 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).