Kolten Conrad

Answered

2022-07-02

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

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Answer & Explanation

iskakanjulc

Expert

2022-07-03Added 18 answers

Notice, we have

${\mathrm{cos}}^{2}\theta -\mathrm{sin}\theta =1$

$1-{\mathrm{sin}}^{2}\theta -\mathrm{sin}\theta =1$

$-{\mathrm{sin}}^{2}\theta -\mathrm{sin}\theta =0$

${\mathrm{sin}}^{2}\theta +\mathrm{sin}\theta =0$

$\mathrm{sin}\theta (\mathrm{sin}\theta +1)=0$

$\mathrm{sin}\theta =0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\theta =n({180}^{\circ})$

Where, n is any integer.

But for the given interval $[{0}^{\circ},{360}^{\circ}]$, substituting n=0,1,2, we get

$\theta ={0}^{\circ},{180}^{\circ},{360}^{\circ}$

Now,

$\mathrm{sin}\theta +1=0$

$\mathrm{sin}\theta =-1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\theta =2n({180}^{\circ})-{90}^{\circ}$

But for given interval $[{0}^{\circ},{360}^{\circ}]$, substituting n=1 we get

$\theta ={270}^{\circ}$

Hence, we have

${\theta}=\left\{{{0}^{\circ},{180}^{\circ},{270}^{\circ},{360}^{\circ}}\right\}$

2d3vljtq

Expert

2022-07-04Added 5 answers

${\mathrm{cos}}^{2}(\theta )-\mathrm{sin}(\theta )={\mathrm{cos}}^{2}(\theta )+{\mathrm{sin}}^{2}(\theta )$ , (using Pythagorean Identity)

${\mathrm{sin}}^{2}(\theta )+\mathrm{sin}(\theta )=0$

let $x=\mathrm{sin}(\theta )$

${x}^{2}+x=0$

$x=\frac{-1\pm \sqrt{{1}^{2}-4(1)(0)}}{2}$, (Quadratic Formula)

x=0, x=−1

substitute $\mathrm{sin}(\theta )$ back in and solve

$0=\mathrm{sin}(\theta )\Rightarrow \theta ={0}^{\circ},{180}^{\circ},{360}^{\circ}$

$-1=\mathrm{sin}(\theta )\Rightarrow \theta =-{270}^{\circ}$

$\theta =\{{0}^{\circ},{180}^{\circ},{270}^{\circ},{360}^{\circ}\}$

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