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Solve the Following Equation: 3 – 2 Cos X – 4 Sin X – Cos 2x + Sin 2x = 0 - Mathematics

Sum

Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0

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Solution

\[3 - 2 \cos x - 4 \sin x - \cos 2x + \sin 2x = 0\]
\[ \Rightarrow 3 - 2 \cos x - 4 \sin x - \left( 1 - 2 \sin^2 x \right) + 2 \sin x \cos x = 0\]
\[ \Rightarrow 3 - 2 \cos x - 4 \sin x - 1 + 2 \sin^2 x + 2 \sin x \cos x = 0\]
\[ \Rightarrow \left( 2 \sin^2 x - 4 \sin x + 2 \right) + 2 \cos x\left( \sin x - 1 \right) = 0\]
\[ \Rightarrow 2\left( \sin^2 x - 2 \sin x + 1 \right) + 2 \cos x\left( \sin x - 1 \right) = 0\]
\[ \Rightarrow 2 \left( \sin x - 1 \right)^2 + 2 \cos x\left( \sin x - 1 \right) = 0\]
\[ \Rightarrow \left( \sin x - 1 \right)\left( 2 \sin x - 2 + 2 \cos x \right) = 0\]
\[ \Rightarrow 2\left( \sin x - 1 \right)\left( \sin x + \cos x - 1 \right) = 0\]
\[ \Rightarrow \left( \sin x - 1 \right) = 0\text{ or }\left( \sin x + \cos x - 1 \right) = 0\]
\[ \Rightarrow \sin x = 1\text{ or }\sin x + \cos x = 1\]
\[ \Rightarrow \sin x = \sin\frac{\pi}{2}\text{ or }\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \sin x = \sin\frac{\pi}{2}\text{ or }\sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4}\cos x = \cos \frac{\pi}{4}\]
\[ \Rightarrow \sin x = \sin\frac{\pi}{2}\text{ or }\cos \left( x - \frac{\pi}{4} \right) = \cos \frac{\pi}{4}\]
\[ \Rightarrow x = n\pi + \left( - 1 \right)^n \frac{\pi}{2}\text{ or }x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z}\]
\[ \Rightarrow x = n\pi + \left( - 1 \right)^n \frac{\pi}{2}\text{ or }x = 2n\pi + \frac{\pi}{2}\text{ or }x = 2n\pi, n \in \mathbb{Z}\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 8 | Page 22
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