Question

Solve the equation on the interval
`0 \leq Theta \ \textless \ sin(theta)`

The equation is 2sin^2(theta)=sin(theta)

Answer 1

`\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} (\pi )/(6)\\\\ (5\pi )/(6) \end{cases}`