# Math Help - Sinister Sine

1. ## 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.

2. 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 $\sin x=\cos x\Rightarrow x=\pi /4$.

3. Hi

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

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

$2 \cos x \\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

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

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

6. 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.

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

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