# Thread: Solving for x with sin

1. ## 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.

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

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

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

3. 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:

$sin2\theta=2cos\theta sin\theta$

4. 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

5. the graph looks like this, can anyone interpret this? are the answers above represents the graph?

6. 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 $y = \sqrt{2} \sin{x} + \sin(2x)$

7. 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".

8. Originally Posted by ANDS!
Both graphs are right. You just have a better "window".
you're right ... need to look closer next time.

9. 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

10. Originally Posted by dhiab
Hello : that's the graph of
and now we have three.

11. 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 ?