Question

Find all the solutions in the interval [0,2pi) for 2sin^2x=sinx

Answer 1

`\bf 2sin^2(x)=sin(x)\implies 2sin^2(x)-sin(x)=0 \\\\\\ sin(x)[2sin(x)-1]=0\implies \begin{cases} sin(x)=0\\ \measuredangle x = sin^(-1)(0)\\ \measuredangle x=0\ ,\ \pi \\ ----------\\ 2sin(x)-1=0\\ 2sin(x)=1\\ sin(x)=(1)/(2)\\ \measuredangle x=sin^(-1)\left( (1)/(2) \right)\\ \measuredangle x =(\pi )/(6)\ ,\ (5\pi )/(6) \end{cases}`