[tex]\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
-------------------------------\\\\
-5sin(x)=-2cos^2(x)+4\implies -5sin(x)=-2[1-sin^2(x)]+4
\\\\\\
-5sin(x)=-2+2sin^2(x)+4\implies -5sin(x)=2sin^2(x)+2
\\\\\\
0=2sin^2(x)+5sin(x)+2\implies [2sin(x)+1][sin(x)+2]=0\\\\
-------------------------------\\\\
sin(x)+2=0\implies sin(x)=-2\impliedby
\begin{array}{llll}
\textit{sine function is never}\\
\textit{more or less than }\pm 1\\
\textit{so, no dice on that one}
\end{array}[/tex]
[tex]\bf 2sin(x)+1=0\implies 2sin(x)=-1\implies sin(x)=-\cfrac{1}{2}
\\\\\\
sin^{-1}[sin(x)]=sin\left( -\frac{1}{2} \right)\implies \measuredangle x=sin\left( -\frac{1}{2} \right)\implies \measuredangle x=
\begin{cases}
\frac{7\pi }{6}\\\\
\frac{11\pi }{6}
\end{cases}[/tex]
so hmm, notice above, it's just a quadratic equation and thus we solve for [sin(x)[, the variable.
on a side note, [tex]\bf \ [cos(\theta )]^n\implies cos^n(\theta )
\\\\\\
\ [sin(\theta )]^n\implies sin^n(\theta )
\\\\\\
\ [tan(\theta )]^n\implies tan^n(\theta )
\\\\\\
so \qquad cos^2(x)\implies [cos(x)]^2[/tex]
just pointing that out, since the exponent on trigonometric functions, can be a bit misleading, and often times you're better off working with the bracketed version, is less ambiguous.
