Mathematics, 20.12.2019 18:31 alananicoleee
Find all the solutions in the interval [0,2pi) for 2sin^2x=sinx
Answers: 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)=\frac{1}{2}\\ \measuredangle x=sin^{-1}\left( \frac{1}{2} \right)\\ \measuredangle x =\frac{\pi }{6}\ ,\ \frac{5\pi }{6} \end{cases}](/tpl/images/0427/9557/34393.png)
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