How do you determine the local extrema points for y=\sqrt{3}\cos(3x)+\sin(3x)

Reginald Owens

Reginald Owens

Answered question

2022-02-26

How do you determine the local extrema points for y=3cos(3x)+sin(3x) ?

Answer & Explanation

Tail3vn

Tail3vn

Beginner2022-02-27Added 3 answers

At x=π2, y0;y(π2)=33
You need to solve for
y=33sin(3x)+3cos(3x)=0
33sin(3x)=cos(3x)
3sin(3x)=cos(3x)3sin(3x)cos(3x)=3tan(3x)=1
tan(3x)=13
lucratifar1

lucratifar1

Beginner2022-02-28Added 5 answers

y=3cos(3x)+sin(3x)=2sin(3x+π3)
y=23cos(3x+π3)
For the extreme values of y,y=0cos(3x+π3)=0
3x+π3=(2n+1)π2 whether n is any integer
3x=(6n+1)π6
As 0x2π3, 03x2π
0(6n+1)π62π06n+112n
=0,1

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