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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.

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 $$16^{sin^2x} = 2iff sin^2x = frac14\81^{cos^2x} = 9iff cos^2x = frac12$$
  Knowing that $sin^2x + cos^2x = 1$, we see that this one is impossible


• 
  $8 + 3$: This time we get
  $$
  16^{sin^2 x} = 8iff sin^2x = frac34\81^{cos^2x} = 3iff cos^2x = frac14
  $$
  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.

Let us write $t:=3^{4sin^2x}$ and $a:=log_36$. Then the equation may be written$$t^a-11t+81=0qquad(1)$$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$$x=npipmtfrac12arccos(1-tfrac12log_3t)quad(ninBbb Z).$$

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

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 $frac{log(16)16}{log(81)} geq (16times 81)^{(cos x)^2}$; this is the case if and only if $frac{log (frac{log(16)16}{log(81)})}{log (16times 81)} geq (cos x)^2$

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 $frac{log (frac{log(16)16}{log(81)})}{log (16times 81)}$. $a$ 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 $[0,2pi]$ 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 $2pi$)

You are looking for the zero of
$$f(y)=16^y+81^{1-y}-11qquad text{where} qquad y=sin^2(x)$$
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
$$f(y)simeq f(y_*)+frac 12 f''(y_*)(y-y_*)^2implies y_pm=y_* pm sqrt{-frac {2f(y_*)}{f''(y_*) } }$$ 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 $x$.