Answer
$$A=\frac{3 \sqrt{3}}{4}
\approx 1.299$$
Also, see Fig.(1)
Work Step by Step
Given $$
f(x)=\sin x, \quad g(x)=\cos 2 x, \quad-\frac{\pi}{2} \leq x \leq \frac{\pi}{6}
$$
So, the intersection points can be obtained
by equating the two function $f(x)=g(x)$, therefore we get
$\sin x=\cos 2 x \quad \Rightarrow \sin x=1-2\sin^ 2 x$
$ \Rightarrow 2\sin^ 2 x+\sin x-1=0 \quad \Rightarrow (2 \sin x-1)( \sin x+1)=0$
$ \Rightarrow \sin x=\frac{1}{2}\Rightarrow x=\frac{\pi}{6} \Rightarrow y=\frac{1}{2
}$
and $\sin x=-1\Rightarrow x=-\frac{\pi}{2} \Rightarrow y=-1$
First, we sketch the region bounded by the graphs $f(x)$ and $g(x)$ (see Fig.1)
Secondly, we use the integration to find the area of the region
$$
\begin{aligned} A &=\int_{-\pi / 2}^{\pi / 6}(g( x)-f(x)) d x\\
&=\int_{-\pi / 2}^{\pi / 6}(\cos 2 x-\sin x) d x \\ &=\left[\frac{1}{2} \sin 2 x+\cos x\right]_{-\pi / 2}^{\pi / 6} \\ &=\left(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{2}\right)-(0)\\
&=\frac{3 \sqrt{3}}{4} \\
&\approx 1.299 \end{aligned}
$$
