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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David Clayton

Answered 2021-12-31
Author has **36** answers

Let $u={\mathrm{sin}}^{2}\left(x\right)\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}v={\mathrm{cos}}^{2}\left(x\right)$ , 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 $u\in [0,1]$ such that f(u)=f(1-u). However ${f}^{\prime}\left(u\right)=\mathrm{ln}\left(2\right)(2u+1){2}^{{u}^{2}+u-1}+1>0$ 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 $\mathrm{sin}}^{2}\left(x\right)=u=\frac{12}{$

Define

Donald Cheek

Answered 2022-01-01
Author has **41** answers

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

$[{\mathrm{sin}}^{4}x-{\mathrm{cos}}^{2}x]-[{\mathrm{cos}}^{4}x-{\mathrm{sin}}^{2}x]=-({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x)({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x+1)=-2\mathrm{cos}2x$

Therefore all possible solutions correspond to zeros of the right hand side (which are also automatically zeroes of the left hand side). This gives$x=\pm \frac{\pi}{4}+2\pi \mathbb{Z},\pm \frac{3\pi}{4}+2\pi \mathbb{Z}$

This follows from

Therefore all possible solutions correspond to zeros of the right hand side (which are also automatically zeroes of the left hand side). This gives

nick1337

Answered 2022-01-08
Author has **510** answers

Putting

So we get,

or,

If

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

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

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where

I put the equation into the form