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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

The number of solutions of the equation $\sin 7 \theta - \sin 3 \theta = \sin 4 \theta$ that lie in the interval $(0, \pi)$ is

The number of solutions of the equation $\cos 6x + \cos 4x + \cos 2x = -1$ in the interval $[0, \pi]$ is:

If $1 + \cot \theta = \text{cosec} \theta$,then the general value of $\theta$ is

The general solution of $|\sin x|=\cos x$ is (when $n \in I$) given by

The number of distinct solutions of $\sec \theta + \tan \theta = \sqrt{3}$ for $0 \leqslant \theta \leqslant 2\pi$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module