Question

Find the solution of tanxsecx+secx-2tanx-2=0 on[0,2π)

Answer 1

You can solve this using a combination of double angle identities and harmonic form... right after you've multiplied through by something to make the equation a bit nicer to work with...

Answer 2

sec x (tan x+1)-2 (tanx+1)=0
(tanx +1)(sec x-2)=0
Either tan x+1=0
\[\tan x=-1=-\tan \frac{ \pi }{4 }=\tan \left( \pi-\frac{ \pi }{ 4 } \right),\tan \left( 2\pi-\frac{ \pi }{ 4 } \right)\]
\[x=\frac{3\pi }{ 4 },\frac{ 7\pi }{ 4 } \]
\[\sec x-2=0,\sec x=2,\frac{ 1 }{ \cos x }=2,\cos x=\frac{ 1 }{ 2 }=\cos \frac{ \pi }{ 3 } ,\cos \left( 2\pi-\frac{ \pi }{ 3 } \right)\]
x=?