General Solutions of Simple Trigonometric Equations

General Solutions of Simple Trigonometric Equations

Theorem :- General solution of \( \sin \theta = 0 \Rightarrow \theta = n\pi ,\,\,n \in \rlap{--} z \)\( \sin \theta = 0 \Rightarrow \theta = n\pi ,\,\,n \in \rlap{--} z \)
Proof :- We know that \(\sin 0 = \sin \pi = \sin 2\pi = \sin ( - \pi ) = \sin ( - 2\pi ) = ........ = 0 \)\( \sin 0 = \sin \pi = \sin 2\pi = \sin ( - \pi ) = \sin ( - 2\pi ) = ........ = 0 \)
\( \Rightarrow \sin \theta = 0 \) \( \Rightarrow \sin \theta = 0 \) for all Integral multiples of \(\pi\)
\( \therefore \sin \theta = 0 \Leftrightarrow \theta = 0 \pm \pi \pm 2\pi ,........\, \)
\( \Leftrightarrow \theta = n\pi ,n \in \rlap{--} z \)\( \pi \)
 \( \therefore \sin \theta = 0 \Leftrightarrow \theta = 0 \pm \pi \pm 2\pi ,........\, \)
\( \Leftrightarrow \theta = n\pi ,n \in \rlap{--} z \)General Solution of \( \sin \theta = 0 \)\( \sin \theta = 0 \) is \( \theta = n\pi ,n \in \rlap{--} z \)\( \theta = n\pi ,n \in \rlap{--} z \)
\( \therefore \theta = \{ n\pi |n \in \rlap{--} z\} \)

Theorem :- General solution of cos\(\theta=0\)\(\theta=0\) is \(\theta=(2n+1) \pi/2,n\in z\)\(\theta=(2n+1)\)\(\pi/2,n\in z\)
Proof :- We know that

\( \Leftrightarrow \theta \)\( \Leftrightarrow \theta \) is odd multiple of \(\pi/2\)\( \pi/2 \)

Theorem : General solution of \( \tan \theta = 0 \)\( \tan \theta = 0 \) is \( \theta = n\pi ,n \in \rlap{--} z \)\( \theta = n\pi ,n \in \rlap{--} z \)
Proof : Since \( \tan \theta = 0 \)
\( \Leftrightarrow \frac{{\sin \theta }} {{\cos \theta }} = 0 \)
\( \Leftrightarrow \sin \theta = 0 \)
\( \Leftrightarrow \theta = n\pi ,n \in \rlap{--} z \)
Note :-
i) Since \( \cos ec\theta \geqslant 1, \) (or) \( \cos ec\theta \leqslant - 1 \) then \( \cos ec\theta = 0\,\, \) has no, solution
ii) Since \( sec\theta \geqslant + 1 \) (or) \( sec\theta \leqslant - 1 \) then \( \sec \theta \)=0 has no solution
iii) General solution of \( \cot \theta = 0 \) is \( \theta = n\pi ,n \in \rlap{--} z \)

Theorem : The general solution of \( \sin \theta = k( - 1 \leqslant k \leqslant 1) \) is \( \theta = n\pi + ( - 1)^n \alpha \) where \( n \in \rlap{--} z \) and  is the principal value of \( \theta \)
Proof : Given \( \sin \theta = k( - 1 \leqslant k \leqslant 1) \) and 
Let \( \sin \theta = k = \sin \alpha \)
Consider \( \sin \theta = \sin \alpha \)
\( \Leftrightarrow \sin \theta - \sin \alpha = 0 \)
\( \Leftrightarrow 2\sin \frac{{\theta - \alpha }} {2}.\cos \frac{{\theta + \alpha }} {2} = 0 \)
\( \Leftrightarrow \) either \( \cos \frac{{\theta + \alpha }} {2} = 0 \) (or) \( \sin \frac{{\theta - \alpha }} {2} = 0 \)

Case i) 
\( \Leftrightarrow \theta + \alpha = (2m + 1)\pi ,m \in \rlap{--} z \)
\( \Leftrightarrow \theta = (2m + 1)\pi - \alpha ,m \in \rlap{--} z \) 
\( \Leftrightarrow \theta = \) any odd multiple of \( \pi - \alpha \).
Case ii) \( \sin \left( {\frac{{\theta - \alpha }} {2}} \right) = 0 \) 
\( \Leftrightarrow \sin \left( {\frac{{\theta - \alpha }} {2}} \right) = \sin m\pi ,\,m \in z \)
\( \Leftrightarrow \frac{{\theta - \alpha }} {2} = m\pi ,m \in z \)
\( \Leftrightarrow \theta - \alpha = 2m\pi \)
\( \Leftrightarrow \theta = 2m\pi + \alpha \)
\( \theta = any\,even\,multple\,of\,\pi + \,from\,case(i)\,and\,case(ii) \)
\( \theta = n\pi + ( - 1)^n \alpha \,\,n \in z \)
\( \therefore \)The general solution of \( \sin \theta = k( - 1 \leqslant k \leqslant 1) \) and \( \alpha \in \left[ { - \frac{\pi } {2},\frac{\pi } {2}} \right] \) is the principal value of is \( \theta = n\pi + ( - 1)^n \alpha ,\,\,\,\,\,n \in z \)

Note-1 :- If ‘n’ is odd i.e., n = 2m + 1 we get the solution given by (i) and if ‘n’ is even i.e.,  n = 2m, we get solution given by (ii) Hence all the general solution includes all possible solutions given by (i) and (ii)    
2. The solution set of \( \sin \theta = k = \sin \alpha ,( - 1 \leqslant k \leqslant 1),\alpha \in \left[ { - \frac{\pi } {2},\frac{\pi } {2}} \right] \) and ‘\( \alpha \)’  is the principal value of \( \theta = n\pi + ( - 1)^n \alpha ,n \in z \).

Theorem: The general solution of \( \cos \theta = k = \cos \alpha ( - 1 \leqslant k \leqslant 1) \) and \( \alpha \in [0,\pi ] \) is the principal value of \(\theta\) is \( \theta = 2n\pi \pm \alpha ,n \in z \)
Proof:- Given \( \cos \theta = k( - 1 \leqslant k \leqslant 1) \) and  is the principal value of \( \theta \)
    Let \( \cos \theta = k = \cos \alpha \)
    Consider \( \cos \theta = \cos \alpha \)
    \( \Leftrightarrow \cos \theta - \cos \alpha = 0 \)
    \( \Leftrightarrow - 2\sin \frac{{\theta + \alpha }} {2}.\sin \frac{{\theta - \alpha }} {2} = 0 \)
    \( \Leftrightarrow \) either \( \sin \frac{{\theta + \alpha }} {2} = 0\,\,(or)\sin \frac{{\theta - \alpha }} {2} = 0 \)
    (i) \( \sin \frac{{\theta + \alpha }} {2} = 0\, \Leftrightarrow \frac{{\theta - \alpha }} {2} = n\pi ,n \in z \)
        \( \Leftrightarrow \theta + \alpha = 2n\pi \Leftrightarrow \theta = 2n\pi - \alpha \)    (1)
    (ii) \( \sin \frac{{\theta - \alpha }} {2} = 0\, \Leftrightarrow \frac{{\theta - \alpha }} {2} = n\pi \) 
         \( \Leftrightarrow \theta = 2n\pi + \alpha \)       (2)
combining (1) and (2) we get the general solution, \( \theta = 2n\pi \pm \alpha ,n \in \rlap{--} z \) and \( \alpha \in [0,\pi ] \)  
\( \therefore \)Solution set of \( \cos \theta = k( - 1 \leqslant k \leqslant 1) \) and \( \alpha \in [0,\pi ] \) is \( \left[ {2n\pi \pm \alpha |n \in \rlap{--} z} \right] \)

Theorem:- The general solution of \( \tan \theta = K(K \in R) \) is \( \theta = n\pi + \alpha ,\,\,\,\,n \in z \) and \( \alpha \in \left( { - \frac{\pi } {2},\frac{\pi } {2}} \right) \) is the principal value of \( \theta \).
Proof : Given \( \tan \theta = K(K \in R) \) and \( \alpha \in \left( { - \frac{\pi } {2},\frac{\pi } {2}} \right) \) is the principal value of \( \theta \).
Let \( \tan \theta = K = \tan \alpha \)
consider \( \tan \theta = \tan \alpha \)
\( \Leftrightarrow \frac{{\sin \theta }} {{\cos \theta }} = \frac{{\sin \alpha }} {{\cos \alpha }} \)
\( \Leftrightarrow \sin \theta \cos \alpha - \cos \theta .\sin \alpha = 0 \)
\( \Leftrightarrow \sin (\theta - \alpha ) = 0 \)
\( \Leftrightarrow \theta - \alpha = n\pi ,n \in z \)
\( \Leftrightarrow \theta = n\pi + \alpha ,n \in z \)
Solution set of \( \tan \theta = \tan \alpha \) is \( \theta = \{ n\pi + \alpha |n \in z\} \) 
Note : General solution of \( \sin \theta = \sin \alpha ,\cos \theta = \cos \alpha \), when solved simultaneously is \( \theta = 2n\pi + \alpha ,n \in z \) and \( \alpha \) is the principal value of \( \theta \) lying between 0 and \( 2\pi \).

Theorem : General solutions of 
    i) \( \sin ^2 \theta = \sin ^2 \alpha \)   (ii)\( \cos ^2 \theta = \cos ^2 \alpha \)        (iii)\( \tan ^2 \theta = \tan ^2 \alpha \)  is \( \theta = n\pi \pm \alpha ,n \in \rlap{--} z \) and ‘\( \alpha \)’is the principal value of \( \theta \).

i) \( \sin ^2 \theta = \sin ^2 \alpha \) 
\( \Leftrightarrow \frac{1} {2}(1 - \cos 2\theta ) = \frac{1} {2}(1 - \cos 2\alpha ) \) 
\( \Leftrightarrow \cos 2\theta = \cos 2\alpha \) 
\( \Leftrightarrow 2\theta = 2n\pi \pm 2\alpha ,n \in z \)
\( \Leftrightarrow \theta = n\pi \pm \alpha ,n \in z \)

ii) \( \cos ^2 \theta = \cos ^2 \alpha \)
\( \Leftrightarrow \frac{1} {2}(1 + \cos 2\theta ) = \frac{1} {2}(1 + \cos 2\alpha ) \)
\( \Leftrightarrow 1 + \cos 2\theta = 1 + \cos 2\alpha \)

\( \Leftrightarrow \cos 2\theta = \cos 2\alpha \)
\( \Leftrightarrow 2\theta = 2n\pi \pm 2\alpha \)
\( \Leftrightarrow \theta = n\pi \pm \alpha ,n \in z \)

iii)\( \tan ^2 \theta = \tan ^2 \alpha \)
\( \Leftrightarrow \frac{1} {{\tan ^2 \theta }} = \frac{1} {{\tan ^2 \alpha }} \)
\( \Leftrightarrow \frac{{1 - \tan ^2 \theta }} {{1 + \tan ^2 \theta }} = \frac{{1 - \tan ^2 \alpha }} {{1 + \tan ^2 \alpha }} \) { Componendo and dividendo}
\( \Leftrightarrow \cos 2\theta = \cos 2\alpha \)
\( \Leftrightarrow 2\theta = 2n\pi \pm 2\alpha ,n \in z \)
\( \Leftrightarrow \theta = n\pi \pm \alpha ,n \in \rlap{--} z \)