Trigonometric Equations and Inequalities
2.0 General Solution of Trigonometric Functions
2.0 General Solution of Trigonometric Functions
1. $\sin \theta = 0$$$\theta = n\pi ,\quad n \in Z$$
Proof:
From the given figure,
$$\sin \theta = \frac{{PM}}{{OP}}$$
$$\begin{equation} \begin{aligned} \therefore \sin \theta = 0 \\ \Rightarrow \frac{{PM}}{{OP}} = 0 \\ \Rightarrow PM = 0 \\ \Rightarrow OP\;coincides\;OX\;and\;OX' \\ \Rightarrow \theta = 0,\;\pi ,\;2\pi , ..., - \pi ,\; - 2\pi ,\; - 3\pi ... \\\end{aligned} \end{equation} $$ Now, we can say that it is forming an Airthmetic Progression (A.P.) with common difference $1$, so the general term is written as $$ \Rightarrow \theta = n\pi ,\;n \in Z\;(Z \to {\text{set of integers}})$$
2. $\tan \theta = 0$$$\Leftrightarrow \;\theta = n\pi ,\quad n \in Z$$
Proof:
From the given figure,
$$\tan \theta = \frac{{PM}}{{OM}}$$
$$\begin{equation} \begin{aligned} \therefore \tan \theta = 0 \\ \Rightarrow \frac{{PM}}{{OM}} = 0 \\ \Rightarrow PM = 0 \\ \Rightarrow OM\;coincides\;OX\;and\;OX' \\ \Rightarrow \theta = 0,\;\pi ,\;2\pi ..., - \pi ,\; - 2\pi ,\; - 3\pi ... \\\end{aligned} \end{equation} $$Now, we can say that it is forming an Airthmetic Progression (A.P.) with common difference $1$, so the general term is written as $$ \Rightarrow \theta = n\pi ,\;n \in Z\;(Z \to {\text{set of integers}})$$
3. $\cos \theta = 0$$$\Leftrightarrow \;\theta = \left( {2n + 1} \right)\frac{\pi }{2},\quad n \in Z$$
Proof:
$$\cos \theta = \frac{{OM}}{{OP}}$$
$$\begin{equation} \begin{aligned} \therefore \cos \theta = 0 \\ \Rightarrow \frac{{OM}}{{OP}} = 0 \\ \Rightarrow OM = 0 \\ \Rightarrow OP\;coincides\;OY\;and\;OY' \\ \Rightarrow \theta = \pm \frac{\pi }{2},\; \pm \frac{{3\pi }}{2},\; \pm \frac{{5\pi }}{2},\; \pm \frac{{7\pi }}{2}... \\\end{aligned} \end{equation} $$ Now, we can say that it is forming an Airthmetic Progression (A.P.) with common difference $1$, so the general term is written as $$ \Rightarrow \theta = \left( {2n + 1} \right)\frac{\pi }{2},\;n \in Z\;\left( {{\text{set of integers}}} \right){\text{ }}$$
4. $\cot \theta = 0$$$\Leftrightarrow \;\theta = \left( {2n + 1} \right)\frac{\pi }{2},\quad n \in Z$$
Proof::
$$\cot \theta = \frac{{OM}}{{PM}}$$
$$\begin{equation} \begin{aligned} \therefore \cot \theta = 0 \\ \Rightarrow \frac{{OM}}{{PM}} = 0 \\ \Rightarrow OM = 0 \\ \Rightarrow PM\;coincides\;OY\;and\;OY' \\ \Rightarrow \theta = \pm \frac{\pi }{2},\; \pm \frac{{3\pi }}{2},\; \pm \frac{{5\pi }}{2},\; \pm \frac{{7\pi }}{2}... \\ \Rightarrow \theta = \left( {2n + 1} \right)\frac{\pi }{2},\;n \in Z \\\end{aligned} \end{equation} $$
NOTE:
- Since, the range of $\sec \theta $ is $$\sec \theta \geqslant 1$$ or $$\sec \theta \leqslant - 1$$ Therefore, $$\sec \theta = 0$$ has no solution.
- Similarly, $$\operatorname{cosec} \theta = 0$$ has no solution.
5. $\sin \theta = \sin \alpha $$$\Leftrightarrow \theta = n\pi + {\left( { - 1} \right)^n}\alpha ,\quad n \in Z$$
Proof: $$\begin{equation} \begin{aligned} \sin \theta = \sin \alpha \\ \Rightarrow \sin \theta - \sin \alpha = 0 \\ \Rightarrow 2\sin \left( {\frac{{\theta - \alpha }}{2}} \right)\cos \left( {\frac{{\theta + \alpha }}{2}} \right) = 0 \\ \Rightarrow \sin \left( {\frac{{\theta - \alpha }}{2}} \right) = 0\quad or,\;\cos \left( {\frac{{\theta + \alpha }}{2}} \right) = 0 \\ \Rightarrow \left( {\frac{{\theta - \alpha }}{2}} \right) = n\pi \quad or,\;\left( {\frac{{\theta + \alpha }}{2}} \right) = \left( {2n + 1} \right)\frac{\pi }{2},\;n \in Z \\ \Rightarrow \theta = 2n\pi + \alpha \quad or,\;\theta = \left( {2n + 1} \right)\pi - \alpha ,\;n \in Z \\ \Rightarrow \theta = (Even\;multiple\;of\;\pi ) + \alpha \\ or,\;\theta = (Odd\;multiple\;of\pi ) - \alpha \\ \Rightarrow \theta = n\pi + {\left( { - 1} \right)^n}\alpha ,\;n \in Z \\\end{aligned} \end{equation} $$
6. $\cos \theta = \cos \alpha $$$\Leftrightarrow \theta = 2n\pi \pm \alpha ,\quad n \in Z$$
Proof:
$$\begin{equation} \begin{aligned} \cos \theta = \cos \alpha \\ \Rightarrow \cos \theta - \cos \alpha = 0 \\ \Rightarrow - 2\sin \left( {\frac{{\theta - \alpha }}{2}} \right)\sin \left( {\frac{{\theta + \alpha }}{2}} \right) = 0 \\ \Rightarrow \sin \left( {\frac{{\theta - \alpha }}{2}} \right) = 0\quad or,\;\sin \left( {\frac{{\theta + \alpha }}{2}} \right) = 0 \\ \Rightarrow \left( {\frac{{\theta - \alpha }}{2}} \right) = n\pi \quad or,\;\left( {\frac{{\theta + \alpha }}{2}} \right) = n\pi ,\;n \in Z \\ \Rightarrow \theta = 2n\pi + \alpha \quad or,\;\theta = 2n\pi - \alpha ,\;n \in Z \\ \Rightarrow \theta = 2n\pi \pm \alpha ,\;n \in Z \\\end{aligned} \end{equation} $$
7. $\tan \theta = \tan \alpha $$$\Leftrightarrow \theta = n\pi + \alpha ,\quad n \in Z$$
Proof:
$$\begin{equation} \begin{aligned} \tan \theta = \tan \alpha \\ \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\sin \alpha }}{{\cos \alpha }} \\ \Rightarrow \sin \theta \cos \alpha - \cos \theta \sin \alpha = 0 \\ \Rightarrow \sin \left( {\theta - \alpha } \right) = 0 \\ \Rightarrow \theta - \alpha = n\pi ,\;n \in Z \\ \Rightarrow \theta = n\pi + \alpha ,\;n \in Z \\\end{aligned} \end{equation} $$
NOTE:
- As we know that ${\text{cosec}}\theta {\text{ = }}\frac{1}{{\sin \theta }}$, solution of $$\operatorname{cosec} \theta = \operatorname{cosec} \alpha $$ is same as $$\sin \theta = \sin \alpha $$
- As we know that ${\text{sec}}\theta {\text{ = }}\frac{1}{{\cos \theta }}$, solution of $$sec\theta = \sec \alpha $$ is same as $$\cos \theta = \cos \alpha $$
- As we know that ${\text{cot}}\theta {\text{ = }}\frac{1}{{\tan \theta }}$, solution of $$\cot \theta = \cot \alpha $$ is same as $$\tan \theta = \tan \alpha $$
8. ${\sin ^2}\theta = {\sin ^2}\alpha $$$\Leftrightarrow \theta = n\pi \pm \alpha ,\;n \in Z$$
Proof:
$$\begin{equation} \begin{aligned} {\sin ^2}\theta = {\sin ^2}\alpha \\ \Rightarrow 2{\sin ^2}\theta = 2{\sin ^2}\alpha \quad \quad \quad [Multiply\,2\,on\,both\,sides] \\ \Rightarrow 1 - \cos 2\theta = 1 - \cos 2\alpha \quad \,[use\;\;\cos 2x = 1 - 2{\sin ^2}x] \\ \Rightarrow \cos 2\theta = \cos 2\alpha \\ \Rightarrow 2\theta = 2n\pi \pm 2\alpha \quad \quad \quad \quad [use\quad \cos \theta = \cos \alpha (from\;6)] \\ \Rightarrow \theta = n\pi \pm \alpha, \;n \in Z \\\end{aligned} \end{equation} $$
9. ${\cos ^2}\theta = {\cos ^2}\alpha $$$\Leftrightarrow \theta = n\pi \pm \alpha ,\;n \in Z$$
Proof:
$$\begin{equation} \begin{aligned} {\cos ^2}\theta = {\cos ^2}\alpha \\ \Rightarrow 2{\cos ^2}\theta = 2{\cos ^2}\alpha \quad \quad \quad [Multiply\,2\,on\,both\,sides] \\ \Rightarrow 1 + \cos 2\theta = 1 + \cos 2\alpha \quad \,[use\;\;\cos 2x = 2{\cos ^2}x - 1] \\ \Rightarrow \cos 2\theta = \cos 2\alpha \\ \Rightarrow 2\theta = 2n\pi \pm 2\alpha \quad \quad \quad \quad [use\quad \cos \theta = \cos \alpha (from\;6)] \\ \Rightarrow \theta = n\pi \pm \alpha, \;n \in Z \\\end{aligned} \end{equation} $$
10. ${\tan ^2}\theta = {\tan ^2}\alpha $$$\Leftrightarrow \theta = n\pi \pm \alpha ,\;n \in Z$$
Proof:
$$\begin{equation} \begin{aligned} {\tan ^2}\theta = {\tan ^2}\alpha \\ \Rightarrow {\tan ^2}\theta - {\tan ^2}\alpha = 0 \\ \Rightarrow \left( {\tan \theta - \tan \alpha } \right)\left( {\tan \theta + \tan \alpha } \right) = 0 \\ \Rightarrow \tan \theta - \tan \alpha = 0\quad or,\;\tan \theta + \tan \alpha = 0 \\ \Rightarrow \tan \theta = \tan \alpha \quad or,\;\tan \theta = - \tan \alpha = \tan \left( { - \alpha } \right) \\ \Rightarrow \theta = n\pi + \alpha \quad or,\;\theta = n\pi - \alpha \quad \quad [Using\;(5)] \\ \Rightarrow \theta = n\pi \pm \alpha ,\;n \in Z \\\end{aligned} \end{equation} $$
