Find all solution of the equation in the interval [0,2\pi].

pedzenekO

pedzenekO

Answered question

2021-08-20

Find all solution of the equation in the interval [0,2π].
(cosx1)(sinx+1)=0

Answer & Explanation

Layton

Layton

Skilled2021-08-21Added 89 answers

(cosx1)(sinx+1)=0 
The given equation's solution, i.e., must now be found.
The entire expression will be equal to 0 if any one factor on the left side of the equation is equal to 0.
cosx1=0 
sinx1=0 
Set the first factor equal to 0 
cosx1=0 
cosx=1 
Take the inverse cosine of both sides of the equation to extract x from inside the cosine 
x=arccos(1) 
The exact value of arccos(1) is 0 
x=2π0 
Similarly sin(x)=1 
Taking the inverse sine of both sides of the equation to exact x from inside the sine 
x=arcsin(1) 
The exact value of arcsin(1) is π2 
x=π2 but it should be between [0,2π] 
x=2π+(π2)=3π2 
x=3π2

Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-23Added 2605 answers

Answer is given below (on video)

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