Answer to Question #93164 in Trigonometry for Amey

Question #93164
Sin⁴x-cos^7x=1
1
Expert's answer
2019-08-23T16:27:22-0400

Solution.

Firstly, let us evaluate sin4x\sin^4x and cos27x\cos^27x .

sin4x[0,1]\sin^4x\isin[0,1]

cos27x[0,1]cos^27x\isin[0,1].

Then we can claim that sin4x0\sin^4x\geq0 and cos27x0\cos^27x\geq0 .

Consequently, to get 1 by substraction and sin4x\sin^4x and cos27x\cos^27x , sin4x\sin^4x have to be equal to one and cos27x\cos^27x have to be equal to zero.

Solving the first equation

sin4x=1\sin^4x=1

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

x=π2+πkx=\frac{\pi}{2}+\pi k ,kZk\isin{Z} .

And the second:

cos27x=0cos^27x=0

cos7x=0cos7x=0

7x=π2+πn7x=\frac{\pi}{2}+\pi n

x=π14+πn7x=\frac{\pi}{14}+\frac{\pi n}{7} ,nZn\isin{Z} .

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

If n=3n=3 , x=π2x=\frac{\pi}{2} .

If n=1n=-1 , x=π2x=-\frac{\pi}{2}.

If n=10n=10 , x=3π2x=\frac{3\pi}{2} .

The period of cos7x\cos{7x} is 2π2\pi .We see the pattern of having some of the answers x=π2x=-\frac{\pi}{2} ;x=π2x=\frac{\pi}{2} ;x=3π2x=\frac{3\pi}{2}. The value of cos7x\cos7x

is defined by xx -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 π\pi and we have x=π2+πnx=\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 cos27=0\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 sin4x=1\sin^4x=1 , there is no need to check another values of nn  because there are no possible values of 

nn to increase the intersection .The intersection of two serias of answers cannot be greater than the smaller seria of answers(x=π2+πk)x=\frac{\pi}{2}+\pi k)

x=[π2+πk][π14+πn7]=π2+πkx=[\frac{\pi}{2}+\pi k]\bigcap[\frac{\pi}{14}+\frac{\pi n}{7}]=\frac{\pi}{2}+\pi k

Answer:x=π2+πkx=\frac{\pi}{2}+\pi k .


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