Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation can be written as:
$1-2{\mathit{sin}}^{2}x+ \alpha sinx=2\alpha -7$
$\Rightarrow 2{\mathit{sin}}^{2}x-\alpha \mathit{sin}x+2\alpha -8=0$
$\Rightarrow sinx=\frac{\alpha -4 }{2}, 2 \left(sinx \ne 2\right)$
For at least one solution
$-1\le \frac{\mathrm{\alpha }-4}{2}\mathrm{ }\le 1$
$\Rightarrow \mathrm{\alpha }\in \left[\mathrm{2,6}\right]$