Question

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

Answer 1

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