Sinister Sine

• February 25th 2009, 09:17 AM
Sinister Sine
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.
• February 25th 2009, 09:46 AM
james_bond
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 $\sin x=\cos x\Rightarrow x=\pi /4$.
• February 25th 2009, 09:46 AM
running-gag
Hi

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

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

$2 \cos x \:(\cos x - \sin x) = 0$

A product is equal to 0 if at least one of the terms is equal to 0
• February 25th 2009, 10:15 AM
Ok, so I actually don't need to substitute with t?
• February 25th 2009, 11:32 AM
running-gag
Quote:

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

What is t ?
• February 25th 2009, 11:41 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. :)
• February 25th 2009, 11:50 AM
running-gag
Using t = sin x is not a general rule to solve trig equation (Shake)
• February 25th 2009, 12:04 PM