![\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}](/tpl/images/0294/6401/c83b6.png)
![\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}](/tpl/images/0294/6401/1958c.png)
![\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](/tpl/images/0294/6401/88246.png)
find all solutions for the interval [0, 2pi)
-5sinx=-2cos^2x+4
(note: the cosine is squ...
Mathematics, 06.10.2019 19:00, kaylaw332
find all solutions for the interval [0, 2pi)
-5sinx=-2cos^2x+4
(note: the cosine is squared, but the x is not)
Answers: 2
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