sqrt 2sinx + sin2x=0 for all values of x.

Thank you.

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- Mar 5th 2011, 12:38 AM #1

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- Mar 5th 2011, 12:50 AM #2

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I'll assume that was meant to say 'solve' for all values of x.

We have $\displaystyle \sqrt{2}\sin{x}+\sin{2x} = 0 \Rightarrow \sqrt{2}\sin{x}+2\cos{x}\sin{x} = 0 \Rightarrow \sin{x}\left(\sqrt{2}+2\cos{x}\right) = 0$.

Thus either $\displaystyle \sin{x} = 0$ or $\displaystyle 2\cos{x} = -\sqrt{2} \Rightarrow \cos{x} = -\frac{1}{\sqrt{2}}$. Recall that $\displaystyle \sin{0} = 0 $ and $\displaystyle \cos{\frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

Then we have $\displaystyle \sin{x} = 0 \Rightarrow \sin{x} = \sin{0} \Rightarrow x = 2n\pi$ or $\displaystyle (2n+1)\pi$. We can simply write this as $\displaystyle k\pi $,

where $\displaystyle k\in\mathbb{Z}$ or zero. Also $\displaystyle \cos{x} = -\frac{1}{\sqrt{2}} \Rightarrow \cos{x} = -\cos\frac{\pi}{4} = \cos\left(\pi+\frac{\pi}{4}\right) \Rightarrow x = 2k\pi \pm \frac{5\pi}{4}.$

- Mar 6th 2011, 04:24 AM #3

- Mar 6th 2011, 06:27 AM #4

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You're right, of course.

Cant we also write $\displaystyle cosx = -cos\frac{\pi}{4} \implies cosx = cos(\pi - \frac{\pi}{4})$