2022-07-02

Solving ${\mathrm{cos}}^{2}\theta -\mathrm{sin}\theta =1,\phantom{\rule{1em}{0ex}}\theta \in \left[{0}^{\circ },{360}^{\circ }\right]$

iskakanjulc

Notice, we have
${\mathrm{cos}}^{2}\theta -\mathrm{sin}\theta =1$
$1-{\mathrm{sin}}^{2}\theta -\mathrm{sin}\theta =1$
$-{\mathrm{sin}}^{2}\theta -\mathrm{sin}\theta =0$
${\mathrm{sin}}^{2}\theta +\mathrm{sin}\theta =0$
$\mathrm{sin}\theta \left(\mathrm{sin}\theta +1\right)=0$
$\mathrm{sin}\theta =0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\theta =n\left({180}^{\circ }\right)$
Where, n is any integer.
But for the given interval $\left[{0}^{\circ },{360}^{\circ }\right]$, substituting n=0,1,2, we get
$\theta ={0}^{\circ },{180}^{\circ },{360}^{\circ }$
Now,
$\mathrm{sin}\theta +1=0$
$\mathrm{sin}\theta =-1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\theta =2n\left({180}^{\circ }\right)-{90}^{\circ }$
But for given interval $\left[{0}^{\circ },{360}^{\circ }\right]$, substituting n=1 we get
$\theta ={270}^{\circ }$
Hence, we have
$\theta =\left\{{0}^{\circ },{180}^{\circ },{270}^{\circ },{360}^{\circ }\right\}$

2d3vljtq

${\mathrm{cos}}^{2}\left(\theta \right)-\mathrm{sin}\left(\theta \right)={\mathrm{cos}}^{2}\left(\theta \right)+{\mathrm{sin}}^{2}\left(\theta \right)$ , (using Pythagorean Identity)
${\mathrm{sin}}^{2}\left(\theta \right)+\mathrm{sin}\left(\theta \right)=0$
let $x=\mathrm{sin}\left(\theta \right)$
${x}^{2}+x=0$
$x=\frac{-1±\sqrt{{1}^{2}-4\left(1\right)\left(0\right)}}{2}$, (Quadratic Formula)
x=0, x=−1
substitute $\mathrm{sin}\left(\theta \right)$ back in and solve
$0=\mathrm{sin}\left(\theta \right)⇒\theta ={0}^{\circ },{180}^{\circ },{360}^{\circ }$
$-1=\mathrm{sin}\left(\theta \right)⇒\theta =-{270}^{\circ }$
$\theta =\left\{{0}^{\circ },{180}^{\circ },{270}^{\circ },{360}^{\circ }\right\}$

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