Trigonometric Equations - 2

Trigonometric Equations - 2

Theorem :  General solution of \( a\cos \theta + b\sin \theta = c \) where a,b,c\(\in\)R, \( \left| c \right| \leqslant \sqrt {a^2 + b^2 } \)
Proof : Given equation is \( a\cos \theta + b\sin \theta + c = 0 \)       (1)
    where a,b,c \(\in\)R and \( \left| c \right| \leqslant \sqrt {a^2 + b^2 } \)
    We first reduce the equation (1) to the form cosx = cosy.
    for this put \( a = r\,\cos \alpha ,\,\,b = r\sin \alpha \)
    so that r² = a²+b² 
    and \( \tan \alpha = \frac{b} {a}\,\,\, \Rightarrow \,\,\alpha = \tan ^{ - 1} \frac{b} {a} \)
    Then (1) reduces to 
    \( r\cos \alpha .\cos \theta + r\,\sin \alpha \sin \theta = c \)
    \( \Rightarrow r[\cos \alpha \cos \theta + \sin \alpha .\sin \theta ] \)=c
    \( \Rightarrow r\cos (\alpha - \theta ) = C \)
    \( \Rightarrow \cos (\theta - \alpha ) \)  =\( \frac{{|c|}} {r} = \cos \beta (say) \)
    \( \Rightarrow \theta - \alpha \,\,\,\,\,\,\,\, = 2n\pi \pm \beta ,\,\,n \in z \)
    \( \theta \, = 2n\pi + \alpha \pm \beta ,\,\,n \in z \)

General solution of (1) is
\( \theta = 2n\pi + \alpha \pm \beta ,\,\,n \in z \)  and \( \alpha = \tan ^{ - 1} \frac{b} {a} \)
Note : Which solving trigonometric equation squaring the equation at any step should be avoided as for as possible. If squaring is necessary, check the solution for extraneous values.
2)    Never cancel terms containig unknown terms on the two sides which are in product it may cause loss of genuine solution.
3)    The answer should not contain such values of angles which make any of the terms undefined (or) infinite
4)     Domain should not change, if it changes necessary corrections must be made
5)     Check that denominetor is not zero at any stage while solving equations.
    Some important identities in solving the problems 
    * tanx +tan(120+x) +tan(x-120)                  = 3 tan3x
    * tanx .tan(120+x) tan(120-x)                      = tan3x
    * cosA.cos(60-A).cos(60+A)                          = \( \frac{1} {4}\cos 3A \)
    * sinA.sin(60-A).sin(60+A)                            = \( \frac{1} {4}\sin 3A \)
    *  \( \cos ^3 \alpha + \cos ^3 (120 + \alpha )\cos ^3 (120 - \alpha ) \)               = \( \frac{3} {4}\cos 3\alpha \)  
    * sin³\( \alpha \) +sin³(120-\(\alpha \))+sin³(120+\(\alpha \))             = \( \frac{3} {4}\sin 3\alpha \)