You are here:

# Calculus/Mathematics

Question
Hi, I was hoping you could help me with this question. It was on last-night's homework and I got it wrong. I tried to ask my teacher today, but he is a substitute that wasn't able to explain it very well (my teacher is on maternity leave). If you could show me the answer, as well as how you get that answer, it would be very helpful.

I need to find the exact solutions of sin2x = sinx in the interval [0,2pi]

Thank you.

Hello John,

I think the easiest way is to sue the double angle formula for sin(2x), which
is: sin(2x)=2cos(x)sin(x).  Thus, this gives
2cos(x)sin(x)=sin(x) ==> 2cos(x)sin(x)-sin(x)=0
==> sin(x)[2cos(x)-1]=0
==> sin(x)=0 OR 2cos(x)-1=0
==> sin(x)=0 gives x=0 or x=pi or x=2pi (since you included 2pi in your interval)
2cos(x)-1=0 ==> cos(x)=1/2 which gives x=pi/3 or x=(2/3)pi

OK?

Abe

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment No Comment

Calculus

Volunteer

#### Abe Mantell

##### Expertise

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

##### Experience

Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook