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The general solution of $a\cos x + b\sin x = c,$ where $a,\,\,b,\,\,c$ are constants
$x = n\pi + {\cos ^{ - 1}}\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }}} \right)$
$x = 2n\pi - {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)$
$x = 2n\pi - {\tan ^{ - 1}}\left( {\frac{b}{a}} \right) \pm {\cos ^{ - 1}}\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }}} \right)$
$x = 2n\pi + {\tan ^{ - 1}}\left( {\frac{b}{a}} \right) \pm {\cos ^{ - 1}}\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }}} \right)$
Solution
(d) $\frac{a}{{\sqrt {{a^2} + {b^2}} }}\cos x + \frac{b}{{\sqrt {{a^2} + {b^2}} }}\sin x = \frac{c}{{\sqrt {{a^2} + {b^2}} }}$
$ \Rightarrow $ $\cos \left( {x – {{\cos }^{ – 1}}\frac{a}{{\sqrt {{a^2} + {b^2}} }}} \right) = \frac{c}{{\sqrt {{a^2} + {b^2}} }}$
$ \Rightarrow $ $x – {\cos ^{ – 1}}\frac{a}{{\sqrt {{a^2} + {b^2}} }} = {\cos ^{ – 1}}\frac{c}{{\sqrt {{a^2} + {b^2}} }}$
General solution is,
$x – {\cos ^{ – 1}}\frac{a}{{\sqrt {{a^2} + {b^2}} }} = 2n\pi \pm {\cos ^{ – 1}}\frac{c}{{\sqrt {{a^2} + {b^2}} }}$
or $x = 2n\pi \pm {\cos ^{ – 1}}\frac{c}{{\sqrt {{a^2} + {b^2}} }} + {\cos ^{ – 1}}\frac{a}{{\sqrt {{a^2} + {b^2}} }}$
$x = 2n\pi + {\tan ^{ – 1}}\frac{b}{a} \pm {\cos ^{ – 1}}\frac{c}{{\sqrt {{a^2} + {b^2}} }}$.
Trick : Put $a = b = c = 1$, then
$\cos \left( {x – \frac{\pi }{4}} \right) = \cos \frac{\pi }{4}$
$ \Rightarrow $ $x = 2n\pi + \frac{\pi }{4} \pm \frac{\pi }{4}$
which is given by option $(d).$
