Question

Sol. [Key Concept: The expression ranges in $\lbrack(1 - \sqrt{2)/2,(1 +}\sqrt{2)/2}\rbrack$; only $a = 0,1$ are integers — count the roots of each.]

$$\ sinx(sinx + cosx) \in \left\lbrack \frac{1 - \sqrt{2}}{2},\frac{1 + \sqrt{2}}{2} \right\rbrack $$2 integer will be there $\Rightarrow a = 0,1$

If $a = 0\ sinx(sinx + cosx) = 0$

$\Rightarrow sinx = 0\ $ or $\ sinx + cosx = 0$

$${x = - \pi,0,\pi }{= tanx = - 1}$$

3 solutions

$x = - \frac{\pi}{4},\frac{3\pi}{4}\ 2$ solutions

If $a = 1\ sinx(sinx + cosx) = 1$

$${\sin^{2}x + sinxcosx = 1 }{2\sin^{2}x + 2sinxcosx = 2 }{1 - cos2x + sin2x = 2 }$$$Sin2x - cos2x = 1 \rightarrow$ square

$$Sin4x = 0 $$$4x = - 4\pi, - 3\pi$, $3\pi,4\pi$

$$x = - \pi, - \frac{3\pi}{4}, - \frac{\pi}{2}, - \frac{\pi}{4},0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi $$$\Rightarrow x = - \frac{3\pi}{4}, - \frac{\pi}{2},\frac{\pi}{4},\frac{\pi}{2}\ 4$ solutions

Total 9 solution