The general solution of a cos x + b sin x = c | vedclass

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Question

(D) Divide the equation $a \cos x + b \sin x = c$ by $\sqrt{a^2 + b^2}$:$\frac{a}{\sqrt{a^2 + b^2}} \cos x + \frac{b}{\sqrt{a^2 + b^2}} \sin x = \frac

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) Divide the equation $a \cos x + b \sin x = c$ by $\sqrt{a^2 + b^2}$:$\frac{a}{\sqrt{a^2 + b^2}} \cos x + \frac{b}{\sqrt{a^2 + b^2}} \sin x = \frac{c}{\sqrt{a^2 + b^2}}$Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos \alpha$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin \alpha$,where $	an \alpha = \frac{b}{a}$.Then the equation becomes $\cos(x - \alpha) = \frac{c}{\sqrt{a^2 + b^2}}$.The general solution is $x - \alpha = 2n\pi \pm \cos^{-1} \left( \frac{c}{\sqrt{a^2 + b^2}} ight)$.Substituting $\alpha = 	an^{-1} \left( \frac{b}{a} ight)$,we get:$x = 2n\pi + 	an^{-1} \left( \frac{b}{a} ight) \pm \cos^{-1} \left( \frac{c}{\sqrt{a^2 + b^2}} ight)$.

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

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