![\bf 2sin^2(\theta )=sin(\theta )\implies 2sin^2(\theta )-sin(\theta )=0\implies sin(\theta )[2sin(\theta )-1]=0 \\\\[-0.35em] ~\dotfill\\\\ sin(\theta )=0\implies \theta =sin^{-1}(0)\implies \theta = \begin{cases} 0\\ \pi \\ 2\pi \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 2sin(\theta )-1=0\implies 2sin(\theta )=1\implies sin(\theta )=\cfrac{1}{2} \\\\\\ \theta =sin^{-1}\left( \cfrac{1}{2}\right)\implies \theta = \begin{cases} \frac{\pi }{6}\\\\ \frac{5\pi }{6} \end{cases}](/tpl/images/0368/4669/9dae5.png)
Solve the equation on the interval[tex]0 \leq theta \ \textless \ sin(theta)[/tex]
the e...
Mathematics, 11.11.2019 01:31 miguelperdomo
Solve the equation on the interval)
the equation is 2sin^2(theta)=sin(theta)
Answers: 1
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