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 solutions of $\sin x + \sin 5x = \sin 3x$ in the interval $(0, \frac{\pi}{2})$ are

Find the general solution of $4 \cos 2x - 4 \sqrt{3} \sin 2x + \cos 3x - \sqrt{3} \sin 3x + \cos x - \sqrt{3} \sin x = 0$.

The most general value of $\theta$ satisfying the equations $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is

The solution set of the equation $\sin^2 \theta - \cos \theta = \frac{1}{4}$ in the interval $[0, 2\pi]$ is

The general solution of the trigonometric equation $\tan x + \tan 2x + \tan 3x = \tan x \cdot \tan 2x \cdot \tan 3x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module