Question

Solve the following equation:
3sin²x – 5 sin x cos x + 8 cos² x = 2

Answer 1

\[3 \sin^2 x - 5 \sin x \cos x + 8 \cos^2 x = 2\]
\[ \Rightarrow 3 \sin^2 x - 5 \sin x \cos x + 3 \cos^2 x + 5 \cos^2 x - 2 = 0\]
\[ \Rightarrow 3\left( \sin^2 x + \cos^2 x \right) - 5 \sin x \cos x + 5 \cos^2 x - 2 = 0\]
\[ \Rightarrow 3 - 5 \sin x \cos x + 5 \cos^2 x - 2 = 0\]
\[ \Rightarrow 5 \cos^2 x - 5 \sin x \cos x + 1 = 0\]
\[ \Rightarrow 5\left( 1 - \sin^2 x \right) - 5 \sin x \cos x + 1 = 0\]
\[ \Rightarrow 5 - 5 \sin^2 x - 5 \sin x \cos x + 1 = 0\]
\[ \Rightarrow 5 \sin^2 x + 5 \sin x \cos x - 6 = 0\]
\[\text{ Dividing by }\cos^2 x,\text{ we get }\]
\[ \Rightarrow 5 \tan^2 x + 5 \tan x - 6 \sec^2 x = 0\]
\[ \Rightarrow 5 \tan^2 x + 5 \tan x - 6 - 6 \tan^2 x = 0\]
\[ \Rightarrow - \tan^2 x + 5 \tan x - 6 = 0\]
\[ \Rightarrow \tan^2 x - 5 \tan x + 6 = 0\]
\[ \Rightarrow \tan^2 x - 3 \tan x - 2 \tan x + 6 = 0\]
\[ \Rightarrow \left( \tan x - 3 \right)\left( \tan x - 2 \right) = 0\]
\[ \Rightarrow \left( \tan x - 3 \right) = 0\text{ or }\left( \tan x - 2 \right) = 0\]
\[ \Rightarrow \tan x = 3\text{ or }\tan x = 2\]
\[ \Rightarrow x = n\pi + \tan^{- 1} 3\text{ or }x = n\pi + \tan^{- 1} 2, n \in \mathbb{Z}\]