Question

Find exact solutions on the interval [0, 2pi) of 10sin^2+7sin+1=0

Answer 1

`\bf \stackrel{\stackrel{\textit{notice is just}\qquad ax^2+bx+c=0}{\downarrow }}{10sin^2(\theta )+7sin(\theta )+1}\implies [5sin(\theta )+1][2sin(\theta )+1]=0 \\\\[-0.35em] ~\dotfill\\\\ 5sin(\theta )+1=0\implies 5sin(\theta )=-1\implies sin(\theta )=-\cfrac{1}{5} \\\\\\ \theta =sin^(-1)\left(-\cfrac{1}{5} \right)\implies \theta \approx \begin{cases} 191.54^o\qquad &III~Quadrant\\ 348.46^o\qquad &IV~Quadrant \end{cases} \\\\[-0.35em] ~\dotfill`

`\bf 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} 210^o\qquad &III~Quadrant\\ 330^o&IV~Quadrant \end{cases}`

recall sine is negative on the III and IV Quadrants only.