Question

Sin⁴x-cos^7x=1

Answer 1

Solution.

Firstly, let us evaluate $\sin^4x$ and $\cos^27x$ .

$\sin^4x\isin[0,1]$

$cos^27x\isin[0,1]$.

Then we can claim that $\sin^4x\geq0$ and $\cos^27x\geq0$ .

Consequently, to get 1 by substraction and $\sin^4x$ and $\cos^27x$ , $\sin^4x$ have to be equal to one and $\cos^27x$ have to be equal to zero.

Solving the first equation

$\sin^4x=1$

$\sin{x}=1$ or $\sin{x}=-1$

$x=\frac{\pi}{2}+\pi k$ ,$k\isin{Z}$ .

And the second:

$cos^27x=0$

$cos7x=0$

$7x=\frac{\pi}{2}+\pi n$

$x=\frac{\pi}{14}+\frac{\pi n}{7}$ ,$n\isin{Z}$ .

Then, we are supposed to find the intersection of answers of two equations. By manipulating the value of $n$ we have got next.

If $n=3$ , $x=\frac{\pi}{2}$ .

If $n=-1$ , $x=-\frac{\pi}{2}$.

If $n=10$ , $x=\frac{3\pi}{2}$ .

The period of $\cos{7x}$ is $2\pi$ .We see the pattern of having some of the answers $x=-\frac{\pi}{2}$ ;$x=\frac{\pi}{2}$ ;$x=\frac{3\pi}{2}$. The value of $\cos7x$

is defined by $x$ -intercept of the point on the unit circle. If we consider the abovementioned seria of answers, there are only 2 points included on it, which fit. So that, we can reduce the value of period to $\pi$ and we have $x=\frac{\pi}{2}+\pi n$ .Additionally,we can see a pattern of having matched the answers of second equation to the first. 

It is a part of the seria of answers to the equation $\cos^27=0$

, but not all of them.

Intersection means a set of answers where two separate series match. And since this part of answers matches the seria of answers to the equation $\sin^4x=1$ , there is no need to check another values of $n$  because there are no possible values of 

$n$ to increase the intersection .The intersection of two serias of answers cannot be greater than the smaller seria of answers($x=\frac{\pi}{2}+\pi k)$

$x=[\frac{\pi}{2}+\pi k]\bigcap[\frac{\pi}{14}+\frac{\pi n}{7}]=\frac{\pi}{2}+\pi k$

Answer:$x=\frac{\pi}{2}+\pi k$ .