Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Solve the Following Equation: Cos X + Sin X = Cos 2 X + Sin 2 X - Mathematics

Sum

Solve the following equation:

\[\cos x + \sin x = \cos 2x + \sin 2x\]
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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?
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APPEARS IN

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