Question

Find all solutions of the equation sec2x−2=0.
The answer is A+Bkπ where k is any integer and 0 14 years ago

Answer 1

Find all solutions of the equation sec2x−2=0.
The answer is A+Bkπ where k is any integer and 0<A<π/2
A=? B=?

Answer 2

sec2x = 2

so cos2x = 1/2

so 2x = 2*pi*k +- pi/3

so x = pi*k +- pi/6

since we require that 0 < A < pi/2, then we have to choose the + option

so x = pi*k + pi/6

A = pi/6, B = 1

Answer 3

ans doesnt seem to work for me

Answer 4

I always change sec to 1/cos

\[ \sec^2x−2=0 \] becomes
\[ \frac{1}{\cos^2 x} -2 =0 \]
add 2 to both sides
\[ \frac{1}{\cos^2 x} -2 +2=0 +2\]
simplify
\[ \frac{1}{\cos^2 x} = 2\]
"flip" both sides
\[ \cos^2 x = \frac{1}{2} \]
take the square root of both sides
\[ \cos x = \sqrt{\frac{1}{2} } = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\]

Answer 5

we should write that last line as
\[ \cos x = \pm \frac{\sqrt{2}}{2} \]
this is a well-known result.