Showing that $\alpha$ satisfies the equation $\sin 2x=x$

Question

This is an A level question. For better understanding, I will attach a screenshot of the question and the mark scheme.

Question:

Here's what I have done:

$$A(OBA) = \frac 12r^2α$$ [basic formula for area of sector]

$$A(ONB) = \frac 12r \sin(\frac {BN}{r})$$ [basic formula for area of triangle]

$$2\frac 12r \sin(\frac {BN}{r}) = \frac 12r^2α$$ [because A(ONB) is half A(OBA)]

$$2 \sin(\frac {BN}{r}) = rα$$ [simplifying the equation]

And that's where I'm stuck.

Answer:

Thanks in advance!

Answer 1

The printed answer is absolutely correct. I don't know how the area of triangle happens to be $1\over 2$rsin($BN\over r$) but the area of triangle is $1\over 2$bh (base x height) base = $r\sin(\alpha)$ ; height = $r\cos(\alpha)$