Problem: Find the number of real roots to the equation $$4^{\sin^2{x}}-2^{\cos{2x}}+1=0 \ , \ \ \ \ \ \ \ \ \ 0<x\leq\pi.$$

Attempt: Simplifying i get $$2^{2\sin^2{x}}-2^{\cos{2x}}=2^{2(1-\cos{2x})}-2^{\cos{2x}}=\frac{2^2}{\left(2^{\cos{2x}}\right)^2}-2^{\cos{2x}}=-1.$$

Setting $t=2^{\cos{2x}}$ gives the third degree equation $$\frac{4}{t^2}-t+1=0 \Longleftrightarrow t^3-t^2-4=(t-2)(t^2+t+2)=0 \Longleftrightarrow t_{\mathbb{R}}=2.$$


$$2^{\cos{2x}}=t=2=2^1 \Longleftrightarrow \cos{2x}=1 \Longleftrightarrow x=\pi k \ , \ k\in \mathbb{Z}. $$

So, $k=0\Rightarrow x=0$ and $k=1\Rightarrow x=\pi.$ The only solution of $x$ that lies in the interval $(0,\pi]$ is given when $k=1$, thus the equation only has one solution in the given interval.


1) I got the correct answer, but is my reasoning correct? Just wan't to make sure that the correct answer isn't just an accident/coincidence.

2) During the simplification step, I used that $\cos{2\theta} = 1-2\sin^2{\theta} \Leftrightarrow 2\sin^2{\theta}=1-\cos{2\theta}.$ Is there a more efficient way of using any other identity?

3) I factored the equation $t^3-t^2-4 = 0$ using a computer, but how can I solve it without knowing that one root is 2, in order to factor?

  • 2
    $\begingroup$ $4^{\sin x} = 2^{2\sin x}$ $\endgroup$
    – GAVD
    May 9, 2017 at 10:39
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    $\begingroup$ You've made a mistake in the very first equality: you use $\sin^2(x)=1-\cos(2x)$, while that should've been $\sin^2(x)=\tfrac12(1-\cos(2x))$ $\endgroup$
    – user304329
    May 9, 2017 at 10:53

2 Answers 2


Though you have made a mistake in your first line, but seeing the triviality of the mistake and the fact that you are aware of it now, let us answer your questions one by one.

  1. The reasoning is correct and the answer is certainly not a fluke (for a moment let us shun out your mistake). I am talking generally about the procedural method of finding the answer.

  2. Here is one more efficient way (as far as I am concerned) $$4^{\sin^2{x}}-2^{\cos{2x}}+1=0 \ , \ \ \ \ \ \ \ \ \ 0<x\leq\pi.$$ $$(2^{\sin^2{x}})^2-2^{1-2\sin^{2}{x}}+1=0$$ $$(2^{\sin^2{x}})^2- \frac{2}{(2^{\sin^{2}{x}})^2}+1=0$$ Let us take $(2^{\sin^2{x}})^2=t$ $$\Rightarrow t^2-2+t=0 $$ This can be further solved easily and afterwards, you can put the value of $t$ in the equation and get the answer. Though this is not much simpler, but still I thought of giving this way. Your own method is completely sufficient and right.

3.As far as cubic equations are concerned, there is no single formula(upto high school and undergraduate level) and hence most of the questions that you are going to encounter would require you to check some integral values and will give the right result also. But if you are not much satisfied with my explanation, you can check this website: https://math.vanderbilt.edu/schectex/courses/cubic/

Hope I have covered all the points and there is no doubt left.

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    $\begingroup$ Really good answer, thank you very much! $\endgroup$
    – Parseval
    May 9, 2017 at 12:24

You've made a mistake in the very first equality: you use $\sin^2(x)=1-\cos(2x)$, while that should've been $\sin^2(x)=\tfrac12(1-\cos(2x))$. You've got the right answer; but that's just a happy little accident.

When we fix the first line, we see


Where we can now set $y=2^{\cos{2x}}$ to see $\frac2y-y=-1$. This is just a quadratic, easily solvable: $2-y^2=-y$, resulting in $y=\frac12\pm\frac12\sqrt{1-4\cdot(-2)}$, and since $y=2^{\cos{2x}}$ is to be positive, we must have $y=\tfrac12+\tfrac12\sqrt9=2$. Thus, $2^{\cos{2x}}=2$; this means that $\cos{2x}=1$. Now we're done: $x\in\{k\pi\mid k\in\Bbb Z\}$ so the only number in that interval is $x=\pi$.

So really, there wasn't a third-degree polynomial to factor, just a quadratic. So to answer your three questions:

  1. No, your reasoning was incorrect, the fact you got the right answer was a coincidence.
  2. I don't think other identities would've made this any easier, for this method is simple enough on itself.
  3. You factored $t^3-t^2-4=0$, but you weren't supposed to. However, it is always good to try and see if you can find a root by hand (this doesn't always work, I'd even say this usually doesn't work, but still), so just try out small integers such as $-3$ through $3$ and see if one turns out to be a root. If you find one, you've only got a quadratic left, so from there on it should be easy enough.
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    $\begingroup$ Thanks. However I'd not say that my reasoning was wrong, the only wrong thing was a single number in the usage of the identity. If got $1/2$ instead, my equation in $t$ would have been a quadratic. $\endgroup$
    – Parseval
    May 9, 2017 at 11:44

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