The general solution of acos x + bsin x = c, | vedclass

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Question

(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 $ $\co

Vedclass · Accepted Answer

$x = 2n\pi + {	an ^{ - 1}}\left( {\frac{b}{a}} ight) \pm {\cos ^{ - 1}}\left( {\frac{c}{{\sqrt {{a^2} + {b^2}} }}} ight)$

Vedclass · Answer

(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 $ $\cos \left( {x - {{\cos }^{ - 1}}\frac{a}{{\sqrt {{a^2} + {b^2}} }}} ight) = \frac{c}{{\sqrt {{a^2} + {b^2}} }}$

$ \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 $x = 2n\pi \pm {\cos ^{ - 1}}\frac{c}{{\sqrt {{a^2} + {b^2}} }} + {\cos ^{ - 1}}\frac{a}{{\sqrt {{a^2} + {b^2}} }}$

$x =  2n\pi + {	an ^{ - 1}}\frac{b}{a} \pm {\cos ^{ - 1}}\frac{c}{{\sqrt {{a^2} + {b^2}} }}$.

Trick : Put $a = b = c = 1$, then

$\cos \left( {x - \frac{\pi }{4}} ight) = \cos \frac{\pi }{4}$

$ \Rightarrow $ $x = 2n\pi + \frac{\pi }{4} \pm \frac{\pi }{4}$

which is given by option $(d).$

The general solution of $a\cos x + b\sin x = c,$ where $a,\,\,b,\,\,c$ are constants

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