please help sin cos

**Trigon** · March 5th 2011, 12:38 AM

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

Thank you.

**TheCoffeeMachine** · March 5th 2011, 12:50 AM

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

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

**cupid** · March 6th 2011, 04:24 AM

Originally Posted by TheCoffeeMachine

where $k\in\mathbb{Z}$ or zero.

$\mathbb{Z}$ includes 0

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 $cosx = -cos\frac{\pi}{4} \implies cosx = cos(\pi - \frac{\pi}{4})$

**TheCoffeeMachine** · March 6th 2011, 06:27 AM

Originally Posted by cupid

$\mathbb{Z}$ includes 0

You're right, of course.

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.

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