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.
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.
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:
$\displaystyle sin2\theta=2cos\theta sin\theta$
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