Solving for x with sin

• August 6th 2009, 06:19 PM
Peleus
Solving 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 -

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

Any points in the right direction would be greatly appreciated.
• August 6th 2009, 06:27 PM
malaygoel
Quote:

Originally Posted by Peleus
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 -

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

Any points in the right direction would be greatly appreciated.

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

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

$\sin x(\sqrt{2} + 2 \cos x) = 0$
• August 6th 2009, 08:42 PM
ANDS!
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:

$sin2\theta=2cos\theta sin\theta$
• August 7th 2009, 08:11 AM
dhiab
Quote:

Originally Posted by malaygoel
$\sqrt{2}\sin x + \sin 2x = 0$

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

$\sin x(\sqrt{2} + 2 \cos x) = 0$

Hello : Thank you ...
here is the suite of resolution :

$\sin x\left( {\sqrt 2 + 2\cos x} \right) = 0 \Leftrightarrow \left( {\sin x = 0} \right) \vee \left( {\sqrt 2 + 2\cos x = 0} \right)$
$\sin x = 0 \Leftrightarrow x = n\pi ....(n \in {\rm Z})$

$
\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}
$

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

Thanks(Clapping)
• August 11th 2009, 07:51 AM
pacman
the graph looks like this, can anyone interpret this? are the answers above represents the graph?
• August 11th 2009, 07:57 AM
skeeter
Quote:

Originally Posted by pacman
the graph looks like this, can anyone interpret this? are the answers above represents the graph?

that's not the graph of $y = \sqrt{2} \sin{x} + \sin(2x)$
• August 11th 2009, 08:20 AM
ANDS!
Quote:

Originally Posted by skeeter
that's not the graph of $y = \sqrt{2} \sin{x} + \sin(2x)$

Both graphs are right. You just have a better "window".
• August 11th 2009, 08:50 AM
skeeter
Quote:

Originally Posted by ANDS!
Both graphs are right. You just have a better "window".

you're right ... need to look closer next time.
• August 11th 2009, 09:31 AM
dhiab
Quote:

Originally Posted by pacman
the graph looks like this, can anyone interpret this? are the answers above represents the graph?

Hello : that's the graph of http://www.mathhelpforum.com/math-he...2b4ecdc8-1.gif
• August 11th 2009, 09:32 AM
skeeter
Quote:

Originally Posted by dhiab

and now we have three.
• August 15th 2009, 05:29 PM
thereddevils
Quote:

Originally Posted by skeeter
that's not the graph of $y = \sqrt{2} \sin{x} + \sin(2x)$

what software did you guys use to generate these graphs ?
• August 15th 2009, 05:40 PM
skeeter