Question No. 1: Find the solutions of the following equations which lie in \([0, 2\pi] \)
\((i)\sin x=-\frac{\sqrt{3}}{2} \quad \quad (ii) cosec\,x=2 \quad \quad (iii)sec\,x=-2 \quad \quad (iv)\cot x=\frac{1}{\sqrt{3}} \\ \)
\( \sin x=-\frac{\sqrt{3}}{2} \)
\[\displaystyle \begin{array}{l}\sin x=-\frac{{\sqrt{3}}}{2}\\\text{Sin x is negative in (III) and (IV) quadrant and }\\x={{\sin }^{{-1}}}\left( {-\frac{{\sqrt{3}}}{2}} \right)=-\frac{\pi }{3}\\\text{Therefore}\\x=\pi +\left( {-\frac{\pi }{3}} \right)=\frac{{4\pi }}{3}\quad \text{in III}\\x=2\pi +\left( {-\frac{\pi }{3}} \right)=\frac{{5\pi }}{3}\quad \text{in IV}\end{array}\]
\( cosec\,x=2 \)
\[\displaystyle \begin{array}{l}\csc \theta =2\\\sin \theta =\frac{1}{2}\\\text{Sin }\theta \text{ is positive in I and II quadrant}\\\text{Reference angle:}\\\theta ={{\sin }^{{-1}}}\left( {\frac{1}{2}} \right)=\frac{\pi }{6}\quad \text{in I}\\\theta =\pi -\frac{\pi }{6}=\frac{{5\pi }}{6}\quad \text{in II}\end{array}\]
\( \sec\,x=-2 \)
\[\displaystyle \begin{array}{l}\sec x=-2\\\cos x=-\frac{1}{2}\\\text{cos }x\text{ is negative in II and III quadrant}\\\text{Reference angle}\\x={{\cos }^{{-1}}}\left( {-\frac{1}{2}} \right)=\frac{{2\pi }}{3}\\x=\pi -\left( {\frac{\pi }{3}} \right)=\frac{{2\pi }}{3}\quad \text{in II}\\x=\pi +\left( {\frac{\pi }{3}} \right)=\frac{{4\pi }}{3}\quad \text{in III}\end{array}\]
\( cot x=\frac{1}{\sqrt{3}} \)
\[\displaystyle \begin{array}{l}\cot \theta =\frac{1}{{\sqrt{3}}}\\\tan \theta =\sqrt{3}\\\text{Tan }\theta \text{ is positive in I and III quadrant and }\\\theta ={{\tan }^{{-1}}}\left( {\sqrt{3}} \right)=\frac{\pi }{3}\quad \text{in I}\\\theta =\pi +\frac{\pi }{3}=\frac{{4\pi }}{3}\quad \text{in III}\end{array}\]
Question No.2 Solve the following trigonometric equations:
(i)- \({{\tan }^{2}}\theta =\frac{1}{3} \quad \quad \)(ii)- \( \text{Cose}{{\text{c}}^{\text{2}}}\theta =\frac{4}{3} \quad \quad \)(iii)- \( {{\sec }^{2}}\theta =\frac{4}{3} \quad \quad\)(iv)- \({{\cot }^{2}}\theta =\frac{1}{3} \)
\({{\tan }^{2}}\theta =\frac{1}{3} \)
\[\displaystyle \begin{array}{l}{{\tan }^{2}}\theta =\frac{1}{3}\\\tan \theta =\pm \frac{1}{{\sqrt{3}}}\\\text{When }\tan \theta =\frac{1}{{\sqrt{3}}}\\\theta \text{ is positive in I III quadrant}\\\theta ={{\tan }^{{-1}}}\left( {\frac{1}{{\sqrt{3}}}} \right)=\frac{\pi }{6}\quad \text{in I}\\\theta =\pi +\frac{\pi }{6}=\frac{{7\pi }}{6}\quad \text{in III}\\\text{When }\tan \theta =-\frac{1}{{\sqrt{3}}}\\\tan \theta \text{ is negative in II and IV quadrant}\\\theta =\pi -\frac{\pi }{6}=\frac{{5\pi }}{6}\quad \text{in II}\\\theta =2\pi -\frac{\pi }{6}=\frac{{11\pi }}{6}\quad \text{in IV}\\\text{So }\theta =\left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}\end{array}\]
\( \text{Cose}{{\text{c}}^{\text{2}}}\theta =\frac{4}{3} \)
\[\displaystyle \begin{array}{l}\text{Cose}{{\text{c}}^{\text{2}}}\text{ }\theta =\frac{4}{3}\\\sin \theta =\pm \frac{{\sqrt{3}}}{2}\\\text{When }\sin \theta =\frac{{\sqrt{3}}}{2}\\\text{Sin is positive in Quadrants I and II:}\\\theta ={{\sin }^{{-1}}}\left( {\frac{{\sqrt{3}}}{2}} \right)={{60}^{{^{{}^\circ }}}}\quad \text{in Quadrant I}\\\theta ={{180}^{{^{{}^\circ }}}}-{{60}^{{^{{}^\circ }}}}={{120}^{{^{{}^\circ }}}}\quad \text{in Quadrant II}\\\text{When }\sin \theta \text{ is negative in Quadrants III and IV:}\\\theta ={{\sin }^{{-1}}}\left( {-\frac{{\sqrt{3}}}{2}} \right)=-{{60}^{{^{{}^\circ }}}}\quad \text{in Quadrant IV}\\\theta ={{360}^{{^{{}^\circ }}}}-{{120}^{{^{{}^\circ }}}}={{240}^{{^{{}^\circ }}}}\quad \text{in Quadrant III}\\\text{So }S.S=\left\{ {\frac{\pi }{3},\frac{{2\pi }}{3},-\frac{\pi }{3},\frac{{-2\pi }}{3}} \right\}\end{array}\]
\( {{\sec }^{2}}\theta =\frac{4}{3} \)
\[\displaystyle \begin{array}{l}{{\sec }^{2}}\theta =\frac{4}{3}\\\sec \theta =\pm \frac{2}{{\sqrt{3}}}\\\cos \theta =\pm \frac{{\sqrt{3}}}{2}\\\text{When }\cos \theta \text{ is positive:}\\\cos \theta =\frac{{\sqrt{3}}}{2}\\\text{Cos is positive in Quadrants I and IV:}\\\theta ={{\cos }^{{-1}}}\left( {\frac{{\sqrt{3}}}{2}} \right)={{30}^{{^{{}^\circ }}}}or\frac{\pi }{6}\quad \text{in Quadrant I}\\=2\pi -\frac{\pi }{6}={{330}^{{^{{}^\circ }}}}or\frac{{11\pi }}{6}\quad \text{in Quadrant IV}\\\text{When }\cos \theta \text{ is negative:}\\\theta ={{\cos }^{{-1}}}\left( {-\frac{{\sqrt{3}}}{2}} \right)=\pi -\frac{\pi }{6}=\frac{{5\pi }}{6}\quad \text{in Quadrant II}\\\theta ={{\cos }^{{-1}}}\left( {-\frac{{\sqrt{3}}}{2}} \right)=\pi +\frac{\pi }{6}=\frac{{7\pi }}{6}\quad \text{in Quadrant III}\\\text{So, S}\text{.S = }\left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}\\\end{array}\]
\({{\cot }^{2}}\theta =\frac{1}{3} \)
\[\displaystyle \begin{array}{l}{{\cot }^{2}}\theta =\frac{1}{3}\\\cot ~\theta =\pm \frac{1}{{\sqrt{3}}}\\\tan ~\theta =\pm \sqrt{3}\\\text{When }(\tan ~\theta =\sqrt{3})\\\theta ={{\tan }^{{-1}}}\sqrt{3}=\frac{\pi }{3}\text{ in }I\\\theta =\pi +\frac{\pi }{3}=\frac{{4\pi }}{3}\text{ in }III\\\text{When }(\tan ~\theta =-\sqrt{3})\\\tan ~\theta =-ve\text{ in }II\text{ and }IV\\\theta =\pi -\frac{\pi }{3}=\frac{{2\pi }}{3}\text{ in }II\\\theta =2\pi -\frac{\pi }{3}=\frac{{5\pi }}{3}\text{ in }IV\\\text{So }S.S=\left\{ {\frac{\pi }{3},\frac{{2\pi }}{3},\frac{{4\pi }}{3},\frac{{5\pi }}{3}} \right\}\end{array}\]
Question No. 3 Find the values of \(\theta \) satisfying the equation \(3~{{\tan }^{2}}\theta +2\sqrt{3}\tan ~\theta +1=0\)
\[\displaystyle \begin{array}{l}3~{{\tan }^{2}}\theta +2\sqrt{3}\tan ~\theta +1=0\\{{(\sqrt{3}\tan ~\theta )}^{2}}+2(\sqrt{3}\tan ~\theta )(1)+{{(1)}^{2}}=0\\{{(\sqrt{3}\tan ~\theta +1)}^{2}}=0\\\sqrt{3}\tan ~\theta +1=0\Rightarrow \tan ~\theta =\frac{{-1}}{{\sqrt{3}}}\\\theta ={{\tan }^{{-1}}}\frac{1}{{\sqrt{3}}}=\frac{\pi }{6}\\\theta =\pi -\frac{\pi }{6}=\frac{{5\pi }}{6}\text{ in }II\\\text{or}\\\theta =2\pi -\frac{\pi }{6}=\frac{{11\pi }}{6}\text{ in }IV\end{array}\]
Question No. 4 Find the values of \(\theta \) satisfying the equation \( ta{{n}^{2}}\theta -\sec ~\theta -1=0 \)
\[\displaystyle \begin{array}{l}ta{{n}^{2}}\theta -\sec ~\theta -1=0\\{{\sec }^{2}}\theta -\sec ~\theta -1=0\\(\sec ~\theta -1)(\sec ~\theta +1)-(\sec ~\theta +1)=0\\(\sec ~\theta +1)(\sec ~\theta -1-1)=0\\(\sec ~\theta +1)(\sec ~\theta -2)=0\\\sec ~\theta +1=0\quad \text{or}\quad \sec ~\theta -2=0\\\sec ~\theta =-1\quad \text{or}\quad \sec ~\theta =2\\\cos ~\theta =-1\quad \text{or}\quad \cos ~\theta =\frac{1}{2}\\\theta =\pi \quad \text{or}\quad \theta ={{\cos }^{{-1}}}\frac{1}{2}=\frac{\pi }{3}\text{ in }I\\\cos ~\theta \text{ is +ve in }I\text{ and }IV\text{ quadrant}\\\quad \quad \quad \quad or\quad \theta ={{\cos }^{{-1}}}\frac{1}{2}=\frac{\pi }{3}\text{ in }I\\\quad \quad \quad \quad \text{or}\quad \theta =\frac{\pi }{3}\text{ or }\frac{{5\pi }}{3}\text{ in }IV\\\text{S}\text{.S }=\left\{ {\pi ,\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}\end{array}\]
Question No. 5 Find the values of \(\theta \) satisfying the equation \( 2~\sin \theta +{{\cos }^{2}}\theta -1=0 \)
\[\displaystyle \begin{array}{l}2~\sin \theta +{{\cos }^{2}}\theta -1=0\\2~\sin \theta +1-{{\sin }^{2}}\theta -1=0\\2~\sin \theta -{{\sin }^{2}}\theta =0\\\sin \theta (2-\sin \theta )=0\\\sin \theta =0\quad \text{ or }\quad 2-sin\theta =0\\\theta =0\text{ , }\pi \quad \text{ or }\quad \sin \theta =2\\\quad \quad \quad \quad \text{ or}\quad \text{Not Possible }\\S.S=\left\{ {0,\pi } \right\}\end{array}\]
Question No. 6 Find the values of \(\theta \) satisfying the equation \( 2~{{\sin }^{2}}\theta -sin~\theta =0 \)
\[\displaystyle \begin{array}{l}2~{{\sin }^{2}}\theta -sin~\theta =0\\\sin ~\theta (2\sin ~\theta -1)=0\\\sin ~\theta =0\quad \text{ or }\quad 2sin~\theta -1=0\\\theta =0\text{ , }\pi \quad \text{ or }\quad \sin ~\theta =\frac{1}{2}\\\text{ }\quad \text{ }\quad \quad \quad \text{or }\quad \sin ~\theta \text{ is positive in I and IV quadrant }\\\quad \quad \quad \text{ }\quad \text{ or }\quad \theta ={{\sin }^{{-1}}}\left( {\frac{1}{2}} \right)=\frac{\pi }{6}\text{ in }\to \text{I }\\\quad \quad \quad \text{ }\quad \text{ or }\quad \theta =\pi -\frac{\pi }{6}=\frac{{5\pi }}{6}\text{ in }\to \text{II }\\S.S=\,\left\{ {0\text{,}\pi ,\frac{\pi }{6},\frac{{5\pi }}{6}} \right\}\end{array}\]
Question No. 7 Find the values of \(\theta \) satisfying the equation \( 3~co{{s}^{2}}\theta -2\sqrt{3}sin~\theta ~cos~\theta -3~si{{n}^{2}}\theta =0 \)
\[\displaystyle \begin{array}{l}3~co{{s}^{2}}\theta -2\sqrt{3}sin~\theta ~cos~\theta -3~si{{n}^{2}}\theta =0\\(3~co{{s}^{2}}\theta -2\sqrt{3}sin~\theta ~cos~\theta -3~si{{n}^{2}}\theta =0\\{{(}^{\prime }}{{\div }^{\prime }}by~si{{n}^{2}}\theta )\text{ we get}\\3~co{{t}^{2}}\theta -2\sqrt{3}cot~\theta -3=0\\\text{Subtract and add }\sqrt{\text{3}}\text{cot }~\text{ }\theta \text{ }\\3~co{{t}^{2}}\theta -2\sqrt{3}cot~\theta -\sqrt{3}cot~\theta +\sqrt{3}cot~\theta -3=0\\3~co{{t}^{2}}\theta -3\sqrt{3}cot~\theta +\sqrt{3}cot~\theta -\sqrt{3}\sqrt{3}=0\\3~Cot~\theta (cot~\theta -\sqrt{3})+\sqrt{3}(cot~\theta -\sqrt{3})=0\\(cot~\theta -\sqrt{3})(3~cot~\theta +\sqrt{3})=0\\cot~\theta -\sqrt{3}=0\quad \text{ or}\quad 3~cot~\theta +\sqrt{3}=0\\cot~\theta =\sqrt{3}\quad \text{or}\quad \cot ~\theta =\frac{{-1}}{{\sqrt{3}}}=\frac{{-\sqrt{3}}}{{\sqrt{3}\sqrt{3}}}\\tan~\theta =\frac{1}{{\sqrt{3}}}\quad \text{or}\quad \tan ~\theta =-\sqrt{3}\\tan~\theta \text{ is +ve in I and III}\quad \text{or}\quad \tan ~\theta \text{ is -ve in II and IV }\\\theta =ta{{n}^{{-1}}}\cdot \frac{1}{{\sqrt{3}}}=\frac{\pi }{6}\text{ in I}\quad \text{or}\quad \theta =ta{{n}^{{-1}}}\sqrt{3}=\frac{\pi }{3}\\\theta =\pi +\frac{\pi }{6}=\frac{{7\pi }}{6}\quad \text{in III}\quad \text{or}\quad \theta =\pi -\frac{\pi }{3}=\frac{{2\pi }}{3}\quad \text{in II}\\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{or}\quad \theta =2\pi -\frac{\pi }{3}=\frac{{5\pi }}{3}\quad \text{in IV}\\S.S=\left\{ {\theta =\frac{\pi }{6},\frac{{7\pi }}{6},\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}\end{array}\]
