Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

Solve the Following Equation: Cos X + Sin X = Cos 2 X + Sin 2 X - Mathematics

Sum

Solve the following equation:

$\cos x + \sin x = \cos 2x + \sin 2x$

Solution

$\cos x + \sin x = \cos2x + \sin2x$
$\Rightarrow \cos x - \cos2x = \sin2x - \sin x$
$\Rightarrow - 2 \sin \left( \frac{3x}{2} \right) \sin \left( \frac{- x}{2} \right) = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{3x}{2} \right)$
$\Rightarrow 2 \sin \left( \frac{3x}{2} \right) \sin \left( \frac{x}{2} \right) = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{3x}{2} \right)$
$\Rightarrow 2 \sin \left( \frac{x}{2} \right) \left[ \sin \left( \frac{3x}{2} \right) - \cos \left( \frac{3x}{2} \right) \right] = 0$
$\Rightarrow \sin \frac{x}{2} = 0$ or
$\sin \frac{3x}{2} - \cos \frac{3x}{2} = 0$
$\Rightarrow \sin \frac{x}{2} = \sin 0$ or
$\sin \frac{3x}{2} = \cos \frac{3x}{2}$

⇒ $\frac{x}{2} = n\pi$,

$n \in Z$ or
$\cos \frac{3x}{2} = \cos \left( \frac{\pi}{2} - \frac{3x}{2} \right)$
$\Rightarrow x = 2n\pi, n \in Z$ or
$\frac{3x}{2} = 2m\pi \pm \left( \frac{\pi}{2} - \frac{3x}{2} \right), m \in Z$
⇒ $x = 2n\pi, n \in Z$ or
$\frac{3x}{2} = 2m\pi + \frac{\pi}{2} - \frac{3x}{2}, m \in Z$  (Taking negative sign will give absurd result.)
$x = 2n\pi, n \in Z$ or
$x = \frac{2m\pi}{3} + \frac{\pi}{6}, m \in Z$
Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 4.5 | Page 22