 # I want to solve for x 2^{\sin^4 x-\cos^2 x}-2^{\cos^4 x-\sin^2 x}=\cos Mabel Breault 2021-12-30 Answered
I want to solve for x
${2}^{{\mathrm{sin}}^{4}x-{\mathrm{cos}}^{2}x}-{2}^{{\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x}=\mathrm{cos}2x$
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Let , 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 $f\left(u\right)={2}^{{u}^{2}+u-1}$ ,then we are looking for a $u\in \left[0,1\right]$ such that f(u)=f(1-u). However ${f}^{\prime }\left(u\right)=\mathrm{ln}\left(2\right)\left(2u+1\right){2}^{{u}^{2}+u-1}+1>0$ for all $u\in \left[0,1\right]$. Hence f is injective on [0,1] and thus f(u)=f(1-u) if and only if u=1-u. Thus ${\mathrm{sin}}^{2}\left(x\right)=u=\frac{12}{}$
###### Not exactly what you’re looking for? Donald Cheek
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
$\left[{\mathrm{sin}}^{4}x-{\mathrm{cos}}^{2}x\right]-\left[{\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x\right]=-\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)\left({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x+1\right)=-2\mathrm{cos}2x$
Therefore all possible solutions correspond to zeros of the right hand side (which are also automatically zeroes of the left hand side). This gives $x=±\frac{\pi }{4}+2\pi \mathbb{Z},±\frac{3\pi }{4}+2\pi \mathbb{Z}$
###### Not exactly what you’re looking for? nick1337

Putting ${\mathrm{cos}}^{2}x=a$
${\mathrm{sin}}^{4}x-{\mathrm{cos}}^{2}x=\left(1-a{\right)}^{2}-a={a}^{2}-3a+1$ and ${\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x={a}^{2}-\left(1-a\right)={a}^{2}+a-1$
So we get, ${2}^{{a}^{2}-3a+1}-{2}^{{a}^{2}+a-1}=2a-1$
or, ${2}^{{a}^{2}-3a+1}\left(1-{2}^{4a-2}\right)=2a-1$
If 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
$⇒2{\mathrm{cos}}^{2}x-1=0⇔\mathrm{cos}2x=0⇒2x=\left(2n+1\right)\frac{\pi }{2}$ where n is any integer