cos (sin^-1 1/4 - tan^-1 (-3) )
Note: I posted a picture below with my work and i wanted to see if i'm correct, if not, please help me out, thanks!!! <3
Hello, boltage619!
$\displaystyle \text{Solve over the interval }(0^o,\,360^o)\!: \;10\sin^2\theta - 13\sin\theta - 3 \:=\:0$
Factor: .$\displaystyle (2\sin\theta - 3)(5\sin\theta + 1) \:=\:0$
We have two equations to solve . . .
$\displaystyle 2\sin\theta - 3 \:=\:0 \quad\Rightarrow\quad \sin\theta \:=\:\tfrac{3}{2} \quad\Rightarrow\quad \text{No real roots}$
$\displaystyle 5\sin\theta + 1 \:=\:0 \quad\Rightarrow\quad \sin\theta \:=\:\text{-}\tfrac{1}{5} \quad\Rightarrow\quad \theta \:=\:\sin^{\text{-}1}\left(\text{-}\tfrac{1}{5}\right)$
. . $\displaystyle \theta \:=\:-11.53695903 \quad\Rightarrow\quad \theta \:=\:\begin{Bmatrix}191.5^o \\ 348.5^o\end{Bmatrix}$