• Mar 5th 2011, 12:38 AM
Trigon
sqrt 2sinx + sin2x=0 for all values of x.

Thank you.
• Mar 5th 2011, 12:50 AM
TheCoffeeMachine
I'll assume that was meant to say 'solve' for all values of x.

Quote:

Originally Posted by Trigon
sqrt 2sinx + sin2x=0 for all values of x.

We have $\sqrt{2}\sin{x}+\sin{2x} = 0 \Rightarrow \sqrt{2}\sin{x}+2\cos{x}\sin{x} = 0 \Rightarrow \sin{x}\left(\sqrt{2}+2\cos{x}\right) = 0$.
Thus either $\sin{x} = 0$ or $2\cos{x} = -\sqrt{2} \Rightarrow \cos{x} = -\frac{1}{\sqrt{2}}$. Recall that $\sin{0} = 0$ and $\cos{\frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
Then we have $\sin{x} = 0 \Rightarrow \sin{x} = \sin{0} \Rightarrow x = 2n\pi$ or $(2n+1)\pi$. We can simply write this as $k\pi$,
where $k\in\mathbb{Z}$ or zero. Also $\cos{x} = -\frac{1}{\sqrt{2}} \Rightarrow \cos{x} = -\cos\frac{\pi}{4} = \cos\left(\pi+\frac{\pi}{4}\right) \Rightarrow x = 2k\pi \pm \frac{5\pi}{4}.$
• Mar 6th 2011, 04:24 AM
cupid
Quote:

Originally Posted by TheCoffeeMachine
where $k\in\mathbb{Z}$ or zero.

$\mathbb{Z}$ includes 0

Quote:

Originally Posted by TheCoffeeMachine
Also $\cos{x} = -\frac{1}{\sqrt{2}} \Rightarrow \cos{x} = -\cos\frac{\pi}{4} = \cos\left(\pi+\frac{\pi}{4}\right) \Rightarrow x = 2k\pi \pm \frac{5\pi}{4}.$

Cant we also write $cosx = -cos\frac{\pi}{4} \implies cosx = cos(\pi - \frac{\pi}{4})$
• Mar 6th 2011, 06:27 AM
TheCoffeeMachine
Quote:

Originally Posted by cupid
$\mathbb{Z}$ includes 0

You're right, of course.
Quote:

Cant we also write $cosx = -cos\frac{\pi}{4} \implies cosx = cos(\pi - \frac{\pi}{4})$
We can. But the aggregate values given by $2n\pi\pm\frac{5\pi}{4}$ and $2n\pi\pm\frac{3\pi}{4}$ are the same, so it doesn't matter which one we write.