Find the sationary points of the curve and their nature

crealolobk

crealolobk

Answered question

2021-12-31

Find the sationary points of the curve and their nature for the equation y=excosx for 0xπ2
I derived it and got ex(sinx+cosx)=0
ex has no solution but I don't know how to find the x such that sinx+cosx=0

Answer & Explanation

boronganfh

boronganfh

Beginner2022-01-01Added 33 answers

Hint:
sinx+cosx=0
If and only if
tanx=1
aquariump9

aquariump9

Beginner2022-01-02Added 40 answers

Since you are asking for 0xπ2 it's quite trivial that x=π4 is the only solution of sinx=cosx
Edit:
Reasoning in 0xπ2: notice that cosx=0 only if x=π2 and for x=π2 we have sinπ2=10=cosπ2
So we can say for sure that x=π2 is not a solution of the equation sinx=cosx
Divide by cos(x) both sides of sinx=cosx and you get tanx=1 which is satiesfied in 0xπ2 only for x=π4
Vasquez

Vasquez

Expert2022-01-08Added 669 answers

For all x, cos(x+π2)=sin(x) hence the equation is equivalent to
cos(x)+cos(x+π2)=0
But we also have the identity
cosp+cosq=2cosp+q2cospq2
So your equation is also equivalent to
2cos(x+π4)cosπ4=0
That is, cos(x+π4)=0
But cost=0 if t=π2+kπ,kZ Therefore, your equation is also equivalent to
x=π4+kπ
And the only value of x in [0,π2] occurs for k=0.

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