Find exact solutions algebraically?

**iluvmathbutitshard** · December 1st 2009, 09:40 AM

How do I find the exact solutions algebraically for the following equation?
sin 2x sin x = cos x

Any help is appreciated.

**Jameson** · December 1st 2009, 09:44 AM

Originally Posted by iluvmathbutitshard

How do I find the exact solutions algebraically for the following equation?
sin 2x sin x = cos x

Any help is appreciated.

sin(2x)=2sin(x)cos(x). That should simplify to something easier to solve.

**iluvmathbutitshard** · December 1st 2009, 09:53 AM

Ok, so how would I simplify further?

**Jameson** · December 1st 2009, 09:57 AM

Originally Posted by iluvmathbutitshard

Ok, so how would I simplify further?

Substitute for sin(2x) in your problem. cos(x) will cancel, divide by 2 and you'll get something simpler. I don't understand where you are getting stuck...

**iluvmathbutitshard** · December 1st 2009, 10:04 AM

Ok, thank you. I got it now.

**e^(i*pi)** · December 1st 2009, 10:22 AM

Originally Posted by Jameson

Substitute for sin(2x) in your problem. cos(x) will cancel, divide by 2 and you'll get something simpler. I don't understand where you are getting stuck...

Without knowing the domain you could potentially be dividing by 0 when cos(x)=0 if you cancel. It is better to take cos(x) from both sides and factor out cos(x)

$2sin^2(x)cos(x)-cos(x)=0$

$<br /> cos(x)(2sin^2(x)-1)=0$

Either $cos(x)=0$ or $2sin^2(x)-1=0$

This should give 3 principal solutions and there may be more or less depending on the domain (usually $0 \leq x \leq 2\pi$ )

**Jameson** · December 1st 2009, 10:28 AM

Very good observation. Thank you for pointing this out.

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