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David, Post-Doctoral Degree
Category: Pre-Calculus
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# find all solutions in the interval [0,2pi) sin 2x sin x-cos

### Resolved Question:

find all solutions in the interval [0,2pi)
sin 2x sin x-cos x = 0
x=?
Submitted: 5 years ago.
Category: Pre-Calculus
Expert:  David replied 5 years ago.

David :

hello

David :

give me a minute and I'll post the solution

Customer:

ok no worries.

Customer:

?

David :

sin 2x sin x-cos x = 0
2sinx cosx sin x-cos x = 0
cosx (2(sinx)^2 -1)=0
so you have two options:
1) cos x=0 ==> x= Pi/2 or 3Pi/2
2) 2(sinx)^2 -1=0 ==> sin x = 1/sqrt(2) ==> x=PI/4 or 3Pi/4

David :

hope this helps

David :

actually there are more: with negative

David :

3) 2(sinx)^2 -1=0 ==> sin x = -1/sqrt(2) ==> x=5PI/4 or 7Pi/4

David :

so the complete answer is this:

David :

Answer: x= Pi/2, 3Pi/2,PI/4, 3Pi/4, 5PI/4, 7Pi/4

David :

done

David :

let me know if there are any questions, it will be my pleasure to assist.

Customer:

i have different ones should i stay on this or just go to input a new question

David :

Pleas eopen a new thread and type "for david" I'l pick