How do I find the exact solutions algebraically for the following equation?

sin 2x sin x = cos x

Any help is appreciated.

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- Dec 1st 2009, 09:40 AMiluvmathbutitshardFind exact solutions algebraically?
How do I find the exact solutions algebraically for the following equation?

sin 2x sin x = cos x

Any help is appreciated. - Dec 1st 2009, 09:44 AMJameson
- Dec 1st 2009, 09:53 AMiluvmathbutitshard
Ok, so how would I simplify further?

- Dec 1st 2009, 09:57 AMJameson
- Dec 1st 2009, 10:04 AMiluvmathbutitshard
Ok, thank you. I got it now.

:) - Dec 1st 2009, 10:22 AMe^(i*pi)
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)

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

$\displaystyle

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

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

This should give 3 principal solutions and there may be more or less depending on the domain (usually $\displaystyle 0 \leq x \leq 2\pi $) - Dec 1st 2009, 10:28 AMJameson
Very good observation. Thank you for pointing this out. (Clapping)