Question

1. Prove the identity (sin(2x) / sin x) – (cos(2x) / cos x) = sec x. Please explain steps.

2. Solve 8sin2x – 6 = 0

Answer 1

\[\frac{ \sin 2x }{ \sin x }-\frac{ \cos 2x }{ \cos x }=\frac{ 2 \sin x \cos x }{ \sin x }-\frac{ 2\cos ^2x-1 }{\cos x }=? \]

Answer 2

\[\sin 2x=\frac{ 6 }{ 8 }=\frac{ 3 }{ 4 }\]
\[\frac{ 2 \tan x }{ 1-\tan ^2x }=\frac{ 3 }{ 4 }\]
\[8 \tan x=3-3 \tan ^2 x\]
\[3 \tan ^2 x+8 \tan x-3=0\]
\[3 \tan ^2x+9 \tan x- \tan x-3=0,3 \tan x \left( \tan x+3 \right)-1\left( \tan x+3 \right)=0\]
\[\left( \tan x+3 \right)\left( 3 \tan x-1 \right)=0\]
\[\tan x=-3,\tan x=\frac{ 1 }{ 3 } ,find~x\]