Question

I need help to solve it

Answer 1

$\bf sin(2\theta)-6sin(\theta)=0\\\\ -----------------------------\\\\ \textit{Double Angle Identities} \\ \quad \\ \boxed{sin(2\theta)=2sin(\theta)cos(\theta)} \\ \quad \\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ 2cos^2(\theta)-1 \end{cases} \\ \quad \\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\ -----------------------------$

$\bf 2sin(\theta)cos(\theta)-6sin(\theta)=0\impliedby \textit{common factor} \\\\\\ sin(\theta)[2cos(\theta)-6]=0\implies \begin{cases} sin(\theta)=0\\ \measuredangle \theta=sin^{-1}(0)\\ 0,\pi ,2\pi \\ ----------\\ 2cos(\theta)-6=0\\ cos(\theta)=\frac{6}{2}\\ cos(\theta)=3\\ \measuredangle \theta=cos^{-1}(3)\\ none \end{cases}$

the domain for sine and cosine are eithe -1 or 1 or in between, anything outside of that, like 3, is a value that won't give us an angle