I want to solve this equation \arccos(x)+\arcsin(x^2-x+1)=\frac{\pi}{2} I write this: for all

Irrerbthist6n

Irrerbthist6n

Answered question

2022-01-16

I want to solve this equation
arccos(x)+arcsin(x2x+1)=π2
I write this: for all x[1,1]
arcsin(x2x+1)=π2arccos(x) then x2x+1=sin(π2arccos(x))=cos(arccos(x))=x
x2x+1=xx22x+1=0(x1)2=0x=1
Is it true ?

Answer & Explanation

Lakisha Archer

Lakisha Archer

Beginner2022-01-17Added 39 answers

By definition arccos(t)+arcsin(t)=π2, implying
arccos(x)=π2arcsin(x2x+1)=arccos(x2x+1)
and the arc cosine function, from [1,1]  to  [0,π], is invertible so that this strictly equivalent to
x=x2x+1,    x[1,1]
which has the double root x=1

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