# Exercise (11.5) - Solution

$\displaystyle \begin{array}{l}\text{1}\text{.}\ \ \ \ \ \text{Solve the following equations for 0}{}^\circ \le \theta \le 360{}^\circ .\\\\\ \ \ \ \ \ \ \text{(a)}\ \ \ \ 3\text{cos}\theta -\sin \theta =2\end{array}$

Show/Hide Solution
$\displaystyle \begin{array}{l}\text{(a)}\ \ \ \ 3\text{cos}\theta -\sin \theta =2\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\ \text{cos}\theta -b\sin \theta =c,\\\\\ \ \ \ \ \ \ \ a=3,b=1,c=2\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{3}^{2}}+{{1}^{2}}}}=\sqrt{{10}}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{1}{3}\\\\\therefore \ \ \ \ \ \ \alpha =18{}^\circ 2{6}'\\\\\therefore \ \ \ \ \ \ 3\text{cos}\theta -\sin \theta =2\Rightarrow \sqrt{{10}}\cos (\theta +18{}^\circ 2{6}')=2\\\\\therefore \ \ \ \ \ \ \cos (\theta +18{}^\circ 2{6}')=0.6325\\\\\therefore \ \ \ \ \ \ \theta +18{}^\circ 2{6}'=50{}^\circ 4{6}'\ \ (\text{or)}\ \theta +18{}^\circ 2{6}'=360{}^\circ -50{}^\circ 4{6}'\\\\\therefore \ \ \ \ \ \ \theta =32{}^\circ 2{0}'\ \ (\text{or)}\ \theta =290{}^\circ 4{8}'\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$\displaystyle \ \ \ \ \ \ \ \text{(b)}\ \ \ \ 2\text{sin}\theta -3\cos \theta =3$

Show/Hide Solution
$\displaystyle \begin{array}{l}\text{(b)}\ \ \ \ 2\text{sin}\theta -3\cos \theta =3\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\ \text{sin}\theta -b\text{cos}\theta =c,\\\\\ \ \ \ \ \ \ \ a=2,b=3,c=3\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{2}^{2}}+{{3}^{2}}}}=\sqrt{{13}}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{3}{2}\\\\\therefore \ \ \ \ \ \ \alpha =56{}^\circ 1{9}'\\\\\therefore \ \ \ \ \ \ 2\text{sin}\theta -3\cos \theta =3\Rightarrow \sqrt{{13}}\sin (\theta -56{}^\circ 1{9}')=3\\\\\therefore \ \ \ \ \ \ \sin (\theta -56{}^\circ 1{9}')=0.8321\\\\\therefore \ \ \ \ \ \ \theta -56{}^\circ 1{9}'=56{}^\circ 1{9}'\ \ (\text{or)}\ \theta -56{}^\circ 1{9}'=180{}^\circ -56{}^\circ 1{9}'\\\\\therefore \ \ \ \ \ \ \theta =112{}^\circ 3{8}'\ \ (\text{or)}\ \theta =180{}^\circ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$\displaystyle \ \ \ \ \ \ \ \text{(c)}\ \ \ \ \cos \theta +2\sin \theta =2$

Show/Hide Solution
$\displaystyle \begin{array}{l}\text{(c)}\ \ \ \ \cos \theta +2\sin \theta =2\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\cos \theta +b\sin \theta =c,\\\\\ \ \ \ \ \ \ \ a=1,b=2,c=2\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{1}^{2}}+{{2}^{2}}}}=\sqrt{5}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =2\\\\\therefore \ \ \ \ \ \ \alpha =63{}^\circ 2{6}'\\\\\therefore \ \ \ \ \ \ \cos \theta +2\sin \theta =2\Rightarrow \sqrt{5}\cos (\theta -63{}^\circ 2{6}')=2\\\\\therefore \ \ \ \ \ \ \cos (\theta -63{}^\circ 2{6}')=0.8944\\\\\therefore \ \ \ \ \ \ \theta -63{}^\circ 2{6}'=26{}^\circ 3{4}'\ \ (\text{or)}\ \theta -63{}^\circ 2{6}'=360{}^\circ -26{}^\circ 3{4}'\ \\\\\therefore \ \ \ \ \ \ \theta =90{}^\circ \ \ (\text{or)}\ \theta =396{}^\circ 5{2}'\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$\displaystyle \ \ \ \ \ \ \ \text{(d)}\ \ \ \ 8\text{sin}\theta +6\cos \theta =5$

Show/Hide Solution
$\displaystyle \begin{array}{l}\text{(d)}\ \ \ \ 8\sin \theta +6\cos \theta =5\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\sin \theta +b\cos \theta =c,\\\\\ \ \ \ \ \ \ \ a=8,b=6,c=5\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{8}^{2}}+{{6}^{2}}}}=10\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{3}{4}\\\\\therefore \ \ \ \ \ \ \alpha =36{}^\circ 5{2}'\\\\\therefore \ \ \ \ \ \ 8\sin \theta +6\cos \theta =5\Rightarrow 10\sin (\theta +36{}^\circ 5{2}')=5\\\\\therefore \ \ \ \ \ \ \sin (\theta +36{}^\circ 5{2}')=0.5\\\\\therefore \ \ \ \ \ \ \theta +36{}^\circ 5{2}'=30{}^\circ \ \ (\text{or)}\ \theta +36{}^\circ 5{2}'=150{}^\circ \ (\text{or)}\ \theta +36{}^\circ 5{2}'=390{}^\circ \\\\\therefore \ \ \ \ \ \ \theta =-6{}^\circ 5{2}'\ \ (\text{or)}\ \theta =113{}^\circ {8}'\ (\text{or)}\ \theta =353{}^\circ {8}'\\\\\ \ \ \ \ \ \ \text{Since}\ 0{}^\circ \le \theta \le 360{}^\circ ,\theta =-6{}^\circ 5{2}'\ \text{is impossible}\text{.}\\\\\therefore \ \ \ \ \ \ \theta =113{}^\circ {8}'\ (\text{or)}\ \theta =353{}^\circ {8}'\ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$\displaystyle \ \ \ \ \ \ \ \text{(e)}\ \ \ \ \sqrt{3}\text{cos}\theta +\sin \theta =\sqrt{2}$

Show/Hide Solution
$\displaystyle \begin{array}{l}\text{(e)}\ \ \ \ \sqrt{3}\cos \theta +\sin \theta =\sqrt{2}\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\cos \theta +b\sin \theta =c,\\\\\ \ \ \ \ \ \ \ a=\sqrt{3},b=1,c=\sqrt{2}\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{3+{{1}^{2}}}}=2\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{1}{{\sqrt{3}}}\\\\\therefore \ \ \ \ \ \ \alpha =30{}^\circ \\\\\therefore \ \ \ \ \ \ \sqrt{3}\cos \theta +\sin \theta \Rightarrow 2\cos (\theta -30{}^\circ )=\sqrt{2}\\\\\therefore \ \ \ \ \ \ \cos (\theta -30{}^\circ )=\displaystyle \frac{{\sqrt{2}}}{2}\\\\\therefore \ \ \ \ \ \ \theta -30{}^\circ =45{}^\circ \ \ (\text{or)}\ \theta -30{}^\circ =315{}^\circ \ \\\\\therefore \ \ \ \ \ \ \theta =75{}^\circ \ \ (\text{or)}\ \theta =345{}^\circ \ \ \ \ \ \end{array}$

$\displaystyle \ \ \ \ \ \ \ \text{(f)}\ \ \ \ \text{sin}\theta -\cos \theta =\sqrt{2}$

Show/Hide Solution
$\displaystyle \begin{array}{l}\text{(f)}\ \ \ \ \sin \theta -\cos \theta =\sqrt{2}\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\sin \theta -b\cos \theta =c,\\\\\ \ \ \ \ \ \ \ a=1,b=1,c=\sqrt{2}\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{1}^{2}}+{{1}^{2}}}}=\sqrt{2}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =1\\\\\therefore \ \ \ \ \ \ \alpha =45{}^\circ \\\\\therefore \ \ \ \ \ \ \sin \theta -\cos \theta =\sqrt{2}\Rightarrow \sqrt{2}\sin (\theta -45{}^\circ )=\sqrt{2}\\\\\therefore \ \ \ \ \ \ \sin (\theta -45{}^\circ )=1\\\\\therefore \ \ \ \ \ \ \theta -45{}^\circ =90{}^\circ \ \ \\\\\therefore \ \ \ \ \ \ \theta =135{}^\circ \end{array}$

$\displaystyle \ \ \ \ \ \ \ \text{(g)}\ \ \ \ 4\text{sin}\theta +3\cos \theta =0$

Show/Hide Solution
$\displaystyle \begin{array}{l}\text{(g)}\ \ \ \ 4\sin \theta +3\cos \theta =0\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\sin \theta +b\cos \theta =c,\\\\\ \ \ \ \ \ \ \ a=4,b=3,c=2\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{4}^{2}}+{{3}^{2}}}}=5\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{3}{4}\\\\\therefore \ \ \ \ \ \ \alpha =36{}^\circ 5{2}'\\\\\therefore \ \ \ \ \ \ 4\sin \theta +3\cos \theta =0\Rightarrow 5\sin (\theta +36{}^\circ 5{2}')=0\\\\\therefore \ \ \ \ \ \ \sin (\theta +36{}^\circ 5{2}')=0\\\\\therefore \ \ \ \ \ \ \theta +36{}^\circ 5{2}'=0{}^\circ \ \ \text{(or)}\ \ \ \theta +36{}^\circ 5{2}'=180{}^\circ \ \text{(or)}\ \ \ \theta +36{}^\circ 5{2}'=360{}^\circ \\\\\therefore \ \ \ \ \ \ \theta =-36{}^\circ 5{2}'\ \ \text{(or)}\ \ \ \theta =143{}^\circ {8}'\ \text{(or)}\ \ \ \theta =323{}^\circ {8}'\\\\\ \ \ \ \ \ \ \ \text{Since }0{}^\circ \le \theta \le 360{}^\circ ,\theta =-36{}^\circ 5{2}'\ \text{is impossible}\text{.}\\\\\therefore \ \ \ \ \ \ \theta =143{}^\circ {8}'\ \text{(or)}\ \ \ \theta =323{}^\circ {8}'\end{array}$