Solve y=\frac 13 [\sin x+[\sin x+[\sin x]]] & [y+[y]]=2\cos x \sin x \in (\pi,2\p

pozicijombx

pozicijombx

Answered question

2022-01-23

Solve y=13sinx+[sinx+[sinx]] & [y+[y]]=2cosx
sinx(π,2π)
y=13sinx+[sinx+[sinx]]
y=-1
[1+[1]]=2cosx
cosx=1 which is possible at x=π hence NO SOLUTION sinxπ2
y=13sinx+[sinx+[sinx]]
y=1
[1+[1]]=2cosx
cosx=1 which is possible at x=0 hence NO SOLUTION
Similary if sinx(0,π)π2
y=13[sinx+[sinx+[sinx]]]
y=0
[0+[0]]=2cosx
cosx=0 which is possible at x=π2 hence NO SOLUTION

Answer & Explanation

Damian Roberts

Damian Roberts

Beginner2022-01-24Added 14 answers

I believe that your answer is correct. [x+k]=[x]+k if k is an integer, so the given pair of equations is just equivalent to y=[sinx]  and  [y]=cosx
This gives cosx=[[sinx]]=[sinx] which has no solution.

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