Question

Answer 1

=>(sinx+4)(sinx-1)=0

=>sinx=1

=>x=pi/2+k2pi

Answer 2

\(\Leftrightarrow\left[{}\begin{matrix}s\text{inx}=1\left(TM\right)\\s\text{inx}=-4\left(lo\text{ại v\text{ì} -4}< -1\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)