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I want to solve for $x$ $$ {{2}^{{{\sin }^{4}}x-{{\cos }^{2}}x}}-{{2}^{{{\cos }^{4}}x-{{\sin }^{2}}x}}=\cos 2x $$ but I don't know how to start. Replacing $\sin x$ or $\cos x$ by $y$ led me nowhere because of the right side.
One of the solutions I've found is $x=\pi/4$ but there could be more solutions though.

Américo Tavares
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Student_G
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4 Answers4

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Let $u=\sin^2(x)$ and $v=\cos^2(x)$, then $2^{u^2-v}-2^{v^2-u} = v-u$ and $u+v=1$. Thus $2^{u^2+u-1}+u = 2^{v^2+v-1}+v$.

Define $f(u) = 2^{u^2+u-1}+u$, then we are looking for a $u\in[0,1]$ such that $f(u)=f(1-u)$. However $f^\prime(u) = \ln(2)(2u+1)2^{u^2+u-1}+1>0$ for all $u\in[0,1]$. Hence $f$ is injective on $[0,1]$ and thus $f(u)=f(1-u)$ if and only if $u=1-u$. Thus $\sin^2(x)=u=\frac{1}{2}$.

Abel
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    This is the smartest idea you have! Good job! – NasuSama May 04 '13 at 15:03
  • @NasuSama Thanks! – Abel May 04 '13 at 15:03
  • Thank you for your answer. – Student_G May 04 '13 at 15:15
  • Very nice answer (+1) – user 1591719 May 04 '13 at 15:22
  • I don't understand why $f(u)=f(1-u)$. I substituted 1-u for u, and I didn't get back f(u). – Ovi May 22 '13 at 10:45
  • @Ovi I suppose you're referring to the last line of the answer. I think my phrasing there is throwing you off a bit. I claim that $f(u)=f(1-u)$ if and only if $u=1-u$. – Abel May 22 '13 at 10:48
  • But since $f(u)$ does not equal $f(1-u)$, how can $u=(1-u)$? – Ovi May 22 '13 at 11:22
  • I think you still misunderstand my point. We are looking for a specific $u$ such that $f(u)=f(1-u)$ and I claim that this is only possible if $u=1-u$, since $f$ is injective. – Abel May 22 '13 at 11:24
  • Oh ok I realized where I was confused. I didn't know why specifically you were looking for a u such that @f(u)=f(1-u)@, but now I see that @f(u)=f(v)@ and @v=1-u$. Very nice solution, +1! How did you think of this? – Ovi May 22 '13 at 16:15
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The left side of the equation is greater than $0$ if and only if the right side is less than $0$, and vice versa. This follows from $$ [\sin^4x-\cos^2x]-[\cos^4x-\sin^2x]=-\left(\cos^2x-\sin^2x\right)(\cos^2x+\sin^2x+1)=-2\cos2x.$$ Therefore all possible solutions correspond to zeros of the right hand side (which are also automatically zeroes of the left hand side). This gives $x=\pm\frac{\pi}{4}+2\pi\mathbb{Z},\pm\frac{3\pi}{4}+2\pi\mathbb{Z}$.

Start wearing purple
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0

Putting $\cos^2x=a,$

$\sin^4x-\cos^2x=(1-a)^2-a=a^2-3a+1$ and $\cos^4x-\sin^2x=a^2-(1-a)=a^2+a-1$

So we get, $2^{a^2-3a+1}-2^{a^2+a-1}=2a-1$

or, $2^{a^2-3a+1}(1-2^{4a-2})=2a-1$

If $2a-1>0,$ $4a-2>0\implies 2^{4a-2}>2^0=1\implies $ the left hand side is $<0$

Similarly, if $2a-1<0$ the right hand side is $>0$

If $2a-1=0,$ both sides become $0$

$\implies 2\cos^2x-1=0\iff \cos2x=0\implies 2x=(2n+1)\frac{\pi}2$ where $n$ is any integer

lab bhattacharjee
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We study the injectivity function $f:[0, 1]\rightarrow R$, $f(t)= 2^{t^{2}+t-1}+t$

$f(t)= f(u)$ involving $t=u$ and then $sin^{2}x=cos^{2}x$...

medicu
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