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?
 A: 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.
