Hi all,

Any help here is appreciated, I'm just not sure how to get this problem started.

Solve for all possible values of x -

$\displaystyle \sqrt{2}\sin x + \sin 2x = 0$

Any points in the right direction would be greatly appreciated.

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- Aug 6th 2009, 06:19 PMPeleusSolving for x with sin
Hi all,

Any help here is appreciated, I'm just not sure how to get this problem started.

Solve for all possible values of x -

$\displaystyle \sqrt{2}\sin x + \sin 2x = 0$

Any points in the right direction would be greatly appreciated. - Aug 6th 2009, 06:27 PMmalaygoel
- Aug 6th 2009, 08:42 PMANDS!
Just to add a bit of direction here; remember that most trig problems like this are all testing your ability to recall Trigonometric Identities and relationships. In this case, you are meant to use the Double Angle Identity for sine which says:

$\displaystyle sin2\theta=2cos\theta sin\theta$ - Aug 7th 2009, 08:11 AMdhiab
Hello : Thank you ...

here is the suite of resolution :

$\displaystyle \sin x\left( {\sqrt 2 + 2\cos x} \right) = 0 \Leftrightarrow \left( {\sin x = 0} \right) \vee \left( {\sqrt 2 + 2\cos x = 0} \right)$

$\displaystyle \sin x = 0 \Leftrightarrow x = n\pi ....(n \in {\rm Z})$

$\displaystyle

\begin{array}{l}

\sqrt 2 + 2\cos x = 0 \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2} \\

\cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \cos x = - \cos \frac{\pi }{4} \\

\end{array}

$

$\displaystyle \begin{array}{l}

\cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \cos x = \cos \left( {\pi - \frac{\pi }{4}} \right) \\

\cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \left( {x = \frac{{3\pi }}{4} + 2n\pi } \right)or\left( {x = - \frac{{3\pi }}{4} + 2n\pi } \right)....(n \in {\rm Z}) \\

\end{array}$

Thanks(Clapping) - Aug 11th 2009, 07:51 AMpacman
the graph looks like this, can anyone interpret this? are the answers above represents the graph?

- Aug 11th 2009, 07:57 AMskeeter
- Aug 11th 2009, 08:20 AMANDS!
- Aug 11th 2009, 08:50 AMskeeter
- Aug 11th 2009, 09:31 AMdhiab
**Hello : that's the graph of****http://www.mathhelpforum.com/math-he...2b4ecdc8-1.gif** - Aug 11th 2009, 09:32 AMskeeter
- Aug 15th 2009, 05:29 PMthereddevils
- Aug 15th 2009, 05:40 PMskeeter