Sinister Sine

**Aminekhadir** · February 25th 2009, 08:17 AM

I've tried to solve the equation 2cos^2x= sin 2x for 0 < x < pi

I know I have to express it in either sine or cos. And after that substitute by t and solve the quadratic equation. But I must be doing something wrong, because I keep getting the wrong I answer.
I would appreciate if you could solve and explain to me how it's solved

Thanks a lot,
Amine.

**james_bond** · February 25th 2009, 08:46 AM

Is $2\cos^2x= \sin 2x$ your equation ? If so then $\cos^2x-\cos x\sin x=0$ using the $\sin 2x=2\sin x\cos x$ identity. So you have to solve $\cos x(\cos x-\sin x)=0$ therefore $\cos x=0\Rightarrow x=k\pi \text{ and }\frac 32\pi$ and $\sin x=\cos x\Rightarrow x=\pi /4$ .

**running-gag** · February 25th 2009, 08:46 AM

Hi

$2 \cos^2x= \sin 2x$

$2 \cos^2x= 2 \sin x \cos x$

$2 \cos x \<img src=$ \cos x - \sin x) = 0" alt="2 \cos x \

\cos x - \sin x) = 0" />

A product is equal to 0 if at least one of the terms is equal to 0

**Aminekhadir** · February 25th 2009, 09:15 AM

Ok, so I actually don't need to substitute with t?

**running-gag** · February 25th 2009, 10:32 AM

Originally Posted by Aminekhadir

Ok, so I actually don't need to substitute with t?

What is t ?

**Aminekhadir** · February 25th 2009, 10:41 AM

It's like t = sin x

For example:
3 sin^2(x) = 1 - sin^2(x)
and it becomes
3 t^2 = 1 - t^2

But this is where I get very confused, because we have cos in our equation and not sine? I hope I haven't confused you also. I'm sorry for that.

**running-gag** · February 25th 2009, 10:50 AM

Using t = sin x is not a general rule to solve trig equation

**Aminekhadir** · February 25th 2009, 11:04 AM

Ok, thanks a lot for the help! I think I have confused myself too much by complicating it.

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