How to solve $16^{(\sin x)^2} + 81^{( \cos x)^2} = 11$

Question

How to solve this (for $x$)? $$16^{(\sin x)^2} + 81^{( \cos x)^2} = 11.$$ I tried to write $11$ as $11^1$ and $1 = (\sin x)^2 + (\cos x)^2$ and even some other method to factorize it, but failed in those methods.

Please help me to solve this.

Answer 1

I don't think there is an analytical way to solve this.

That being said, we have some power of $2$ added to some power of $3$ making $11$. Wouldn't it be nice if $2+9$ or $8+3$ were possible?

Let's try!

• $2+9$: We have $$16^{\sin^2x} = 2\iff \sin^2x = \frac14\\81^{\cos^2x} = 9\iff \cos^2x = \frac12$$ Knowing that $\sin^2x + \cos^2x = 1$, we see that this one is impossible
• $8 + 3$: This time we get $$ 16^{\sin^2 x} = 8\iff \sin^2x = \frac34\\81^{\cos^2x} = 3\iff \cos^2x = \frac14 $$ This one is possible. Now we just have to find all such $x$ (which is a relatively standard trigonometry exercise), and we have found the solutions.

Is this the only solution? No, it isn't. WolframAlpha tells us there is another solution at about $\sin^2x \approx 0.6075$, but I have no idea whether it has any nice closed form, or how to arrive at it.

Answer 2

Let's study $f: x\mapsto 16^{(\sin x)^2} + 81^{(\cos x)^2}-11$ defined on $[0,\pi]$. It's a $C^\infty$ function and its derivative is $f'(x) = \log (16)2 \cos (x) \sin(x) 16^{(\sin x)^2} - \log (81)2 \cos (x) \sin(x) 81^{(\cos x)^2} = 2\cos(x)\sin(x)(\log(16)16^{(\sin x)^2} - \log(81)81^{(\cos x)^2})$.

Let's compare $\log(16)16^{(\sin x)^2} - \log(81)81^{(\cos x)^2}$ and $0$. $\log(16)16^{(\sin x)^2} = \log(16) 16 (1/16)^{(\cos x)^2}$ so this quantity is nonnegative if and only if $\frac{\log(16)16}{\log(81)} \geq (16\times 81)^{(\cos x)^2}$; this is the case if and only if $\frac{\log (\frac{\log(16)16}{\log(81)})}{\log (16\times 81)} \geq (\cos x)^2$

According to some calculator I found online, this numerical value on the left is about $0.32259$. This means that this inequality holds for $x$ in an interval centered around $\frac{\pi}{2}$, say $[\frac{\pi}{2} - \epsilon, \frac{\pi}{2}+\epsilon]=[a,b]$ for some $\epsilon > 0$.

So now if we try to draw the sign table for $f'$ we get that on $[0,\frac{\pi}{2}-\epsilon]$, $\cos(x)\sin(x) \geq 0$ and $\log(16)16^{(\sin x)^2} - \log(81)81^{(\cos x)^2}\leq 0$, so $f'(x) \leq 0$.

Similarly, on $[\frac{\pi}{2}-\epsilon, \frac{\pi}{2}]$, $f'(x) \geq 0$, on $[\frac{\pi}{2}, \frac{\pi}{2}+\epsilon]$, $f'(x) \leq 0$ and on $[\frac{\pi}{2}+\epsilon, \pi]$, $f'(x)\geq 0$.

Therefore $f$ is increasing and decreasing on those intervals.

It now remains to see what $f$ is worth on the boundaries to see where it must vanish.

OK so $f(0) = 16^0 + 81^1 - 11 = 71>0$, $f(\pi) = 71>0$, $f(\frac{\pi}{2}) = 6 >0$, so it remains to evaluate $f(a)$ and $f(b)$.

Let's use $p$ to denote $\frac{\log (\frac{\log(16)16}{\log(81)})}{\log (16\times 81)}$. $a$ and $b$ are the two points at which $\cos (x)^2 = p$. Thus $\sin(x)^2 = 1-p$, and so $f(a) = f(b) = 16^{1-p} + 81^p - 11$.

Again according to some calculator found online this is about $-0.33119$, in particular it is $<0$.

Therefore if we draw $f$ we will see that it has $4$ zeroes in $[0,\pi]$ (two around $a$, two around $b$).

You can then find that it must have $8$ zeroes in $[0,2\pi]$ by symmetry; so there are $8$ couples $(\cos(x),\sin(x))$ that satisfy this equation (so infinitely many $x$'s but we can partition the solutions in $8$ classes mod $2\pi$)

Answer 3

Let us write $t:=3^{4\sin^2x}$ and $a:=\log_36$. Then the equation may be written$$t^a-11t+81=0\qquad(1)$$with $t>0$. The upward-curving graph of $y=t^a$ is cut by the straight line $y=11t-81$ just twice. The two roots can be found straightforwardly to any required degree of accuracy by (e.g.) Newton–Raphson. Noting that one of them homes in on an integer value for $\log_3t$, we can test the exact integer and so obtain simply-written solutions for $x$. The other solution (in $t$) has no such closed form, and nor would we expect one, because the exponent $a$ in eqn $1$ is not even rational, let alone belonging to the few rational exponents that would allow a closed-form expression for the solution.

Once you have $t$, you have a general solution$$x=n\pi\pm\tfrac12\arccos(1-\tfrac12\log_3t)\quad(n\in\Bbb Z).$$

Answer 4

You are looking for the zero of $$f(y)=16^y+81^{1-y}-11\qquad \text{where} \qquad y=\sin^2(x)$$ The first derivative $$f'(y)=16^y \log (16)-81^{1-y} \log (81)$$ cancels at $$y_*=\frac{\log (81)-\log \left(\frac{\log (16)}{\log (81)}\right)}{\log (16)+\log (81)}\approx 0.677408$$ and $f(y_*)\approx -0.331191$. The second derivative test shows that this is a minimum; then two roots.

Build the second order Taylor expansion arounf $y_*$. This would give $$f(y)\simeq f(y_*)+\frac 12 f''(y_*)(y-y_*)^2\implies y_\pm=y_* \pm \sqrt{-\frac {2f(y_*)}{f''(y_*) } }$$ leading to $y_- \approx 0.606024$ and $y_+\approx 0.748792$ while the "exact" solutions would be $0.607547$ and $0.750000$.

Now, go back to $x$.