# Find the solutions to ${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }=2$

Find solutions to, $${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }=2$$ where $0≤x≤8π$

Since $\sin x$ and $\cos x$ are in powers hence $450$ and $273$ will never be zero for any $x$. So I took $$(\sin x)^{ 3 }=0$$ and $$(\cos x)^{ 5 }=0$$ and thus I got $5$ solutions in $[0, 8π]$. Is it correct?

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So what should be the correct approach? –  Hyperbola Jan 31 '13 at 4:31
The sine and cosine cannot be simultaneously $0$, so the above is not right. –  André Nicolas Jan 31 '13 at 4:38
@Hyperbola your question is find the number of solutions or find the solutions? –  noname1014 Jan 31 '13 at 4:39
Every number is this equation is weird. –  Patrick Li Jan 31 '13 at 4:43
The original question asked to find the number of solutions only. –  Hyperbola Jan 31 '13 at 4:44

The function ${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }$ is periodic with period $2\pi$, so it suffices to find all solutions in $[0,2\pi)$, then add multiples of $2\pi$ to get the rest.

If $x$ is a solution, then $\sin x$ and $\cos x$ must have opposite signs. This is true because $\sin x$ and $\cos x$ can't both be zero, so if they are both nonnegative then ${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }>2$, while if they are both nonpositive then ${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }<2$. Therefore, all solutions in $[0,2\pi)$ are also in $(\pi/2,\pi)\cup(3\pi/2,2\pi)$.

There is at least one solution in each of the intervals $(\pi/2,\pi)$ and $(3\pi/2,2\pi)$. This can be seen by evaluating ${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }$ when $x=\pi/2, \pi, 3\pi/2, 2\pi$, and noticing that it is respectively $>2,<2,<2,>2$, and applying the Intermediate Value Theorem.

There is at most one solution in each of the intervals $(\pi/2,\pi)$ and $(3\pi/2,2\pi)$. This can be seen by observing that ${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }$ is strictly decreasing on $(\pi/2,\pi)$, and strictly increasing on $(3\pi/2,2\pi)$.

By translating these two solutions by multiples of $2\pi$, there are altogether $8$ solutions in $[0,8\pi]$.

Here is an explanation of why ${ 450 }^{ { (\sin x) }^{ 3 } }+{ 273 }^{ { (\cos x) }^{ 5 } }$ is decreasing on $(\pi/2,\pi)$. If $b>1$, then $b^x$ is an increasing function of $x$. If $k$ is an odd positive integer, then $x^k$ is an increasing function of $x$. On the interval $(\pi/2,\pi)$, $\sin(x)$ is a decreasing function of $x$. The composition of a decreasing function with increasing functions is decreasing, so $$x\mapsto \sin x\mapsto (\sin x)^3\mapsto 450^{(\sin x)^3}$$ is decreasing on $(\pi/2,\pi)$. Similarly, $273^{(\cos x)^5}$ is decreasing on that interval, and a sum of two decreasing functions is decreasing. (I haven't given proofs here, but I've broken it down to simpler statements that do have elementary proofs.) The reason the function is increasing on $(3\pi/2,2\pi)$ is similar; $\sin x$ and $\cos x$ are both increasing on that interval.

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Can you give a quick explanation of the strictly increasing / decreasing part? Did you just take the differential? I got up to there, and then didn't want to differentiate. –  Calvin Lin Jan 31 '13 at 5:08
No, no derivatives needed, I'll add an explanation. –  Jonas Meyer Jan 31 '13 at 5:09
Thanks! That makes sense. I didn't think of the function composition. Though, you'd likely want to get rid of $k$ an (odd positive) integer, since it's likely not an integer, and then just focus on $0 < \sin x < 1$ or $-1 < \sin x < 0$. –  Calvin Lin Jan 31 '13 at 5:19
@Calvin: The values of $k$ I had in mind, relevant to the problem, are $3$ and $5$. I don't understand your comment but that hints to me that I probably wasn't clear that I'm referring to the composition $y=\sin(x)$, $z=y^3$, $w=450^z$. –  Jonas Meyer Jan 31 '13 at 5:22
Ahhhhh ... I was wondering why you switched from $b^x$ to $x^k$. That's much easier to understand now. –  Calvin Lin Jan 31 '13 at 5:25