Question

Find all solutions of the equation sec4x−2=0
The answer is A+k2 and B+k2 where k is any integer and 0 10 years ago

Answer 1

\[\sec (4x)-2=0\]
\[=1/\cos(4x)=2\]
\[=>\cos(4x)=1/2\]
\[4x=\cos^{-1}(1/2)\]
\[4x=\cos^{-1}(0.5)+2k\pi,\ \ k\in \mathbb{Z}\]
\[x=arcCos(0.5)/4 + k·\frac{ \pi }{ 2}, \ \ k\in \mathbb{Z}\]
or X=15º (degree) +k 90º

Answer 2

the reason to have the term 2kpi is because there are many solutions to Cos (x)=0.5 everyone is delayed 360º but i made a mistake, because Cos( X)= cos(-x)=0.5
And de delay is not anymore 360º now it is 180º or "pi"
\[4x=\cos^{-1} (0.5) + k\pi\]
\[X=\cos^{-1}(0.5)/4+k·\pi /4\]

Answer 3

So A=arccos(0.5)/4 and B=PI/4??

Answer 4

they ask you 2 solutions X1=A+K1, X2 =B+K2
then A = B = arccos(0.5)/4
But K1 = when K=1
and K2 = when K =2

Answer 5

if i'm right, the solutions are:
X1=arccos(0.5)/4 + pi/4
X2=arccos(0.5)/4 + pi/2
i don't know what 0<A<B<pi/2 means. i assume typing error
maybe A and B are the solutions
So
A=arccos(0.5)/4+pi/4
B=the same....
you can choose any positive or negative value (integer) of k

Answer 6

Okay, thank you soo much for all your help!