If α be the number of solutions of the equation [sin x]=|x| and β be…

Since, $-1 \leq \sin x \leq 1$ for all $x$. Therefore,

$[\sin x]=\left\{\begin{aligned}
-1, & \text { when }-1 \leq \sin x<0 \\
0, & \text { when } 0 \leq \sin x<1 \\
1, & \text { when } \sin x=1 .
\end{aligned}\right.$

The given equation is

$[\sin x]=|x|$

$\Rightarrow \quad[\sin x] \geq 0$

∴  $[\sin x]=0,1$

Thus, $[\sin x]=|x| \Rightarrow|x|=0,1 \Rightarrow x=0, \pm 1$

But, only $x=0$ satisfies the equation (i). Therefore, $\alpha=1$.

Now,

$\beta$ = Greatest value of $\cos \left(x^2-\left[x^2\right]\right) \text { in }[-1,1]$

$\beta$ = 1                   $\left[∵ \cos \left(x^2-\left[x^2\right]\right)=\cos 0=1 \text { at } x=0\right]$

Hence, $\alpha=\beta$.