Question

3 sin2(θ) + 16 sin(θ) − 35 = 0

Answer 1

`\bf 3sin^2(\theta )+16sin(\theta )-35=0\impliedby \begin{array}{llll} \textit{notice this is just a quadratic }\\\\ax^2+bx+c=0 \end{array} \\\\\\\ [3sin(\theta )-5][sin(\theta )+7]=0\\\\ -------------------------------\\\\ 3sin(\theta )-5=0\implies 3sin(\theta )=5\implies sin(\theta )=\cfrac{5}{3} \\\\\\ \measuredangle \theta =sin^(-1)\left( (5)/(3) \right)`

now, before continuing on the first root, recall that the sine has a range of -1 ⩽ sin(θ) ⩽1, however 5/3 is greater than 1, and therefore is out of range, meaning, there's no such angle, so we've got zilch there.

now, let's check the second root,


`\bf sin(\theta )+7=0\implies sin(\theta )=-7\implies \measuredangle \theta =sin^(-1)(-7)`

well, hell -7 is smaller than -1, so is also outside the range for sine, and so, the second root gave also zilch, so, in short, there's no such angle.