Question

Find all solutions for the interval [0, 2pi)

Answer 1

$\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\ -------------------------------\\\\ -5sin(x)=-2cos^2(x)+4\implies -5sin(x)=-2[1-sin^2(x)]+4 \\\\\\ -5sin(x)=-2+2sin^2(x)+4\implies -5sin(x)=2sin^2(x)+2 \\\\\\ 0=2sin^2(x)+5sin(x)+2\implies [2sin(x)+1][sin(x)+2]=0\\\\ -------------------------------\\\\ sin(x)+2=0\implies sin(x)=-2\impliedby \begin{array}{llll} \textit{sine function is never}\\ \textit{more or less than }\pm 1\\ \textit{so, no dice on that one} \end{array}$

$\bf 2sin(x)+1=0\implies 2sin(x)=-1\implies sin(x)=-\cfrac{1}{2} \\\\\\ sin^{-1}[sin(x)]=sin\left( -\frac{1}{2} \right)\implies \measuredangle x=sin\left( -\frac{1}{2} \right)\implies \measuredangle x= \begin{cases} \frac{7\pi }{6}\\\\ \frac{11\pi }{6} \end{cases}$


so hmm, notice above, it's just a quadratic equation and thus we solve for [sin(x)[, the variable.

on a side note,    $\bf \ [cos(\theta )]^n\implies cos^n(\theta ) \\\\\\ \ [sin(\theta )]^n\implies sin^n(\theta ) \\\\\\ \ [tan(\theta )]^n\implies tan^n(\theta ) \\\\\\ so \qquad cos^2(x)\implies [cos(x)]^2$

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.