Question

Solve the equation 2cosx-secx =1 on the interval [0,2pi).
The only four choices they give me are A. 0, 2pi/3 B. 0, pi/6, 5pi/6 C. 0, 2pi/3, 4pi/9 and D. 0, pi/2, 3pi/2

Answer 1

let \(a=\cos(x)\)
then you have

\(2a-\frac{1}{a}=1\implies2a^2-1=a\implies2a^2-a-1=0\)

now we have a quadratic

\(2a^2-a-1=0\implies (a-1)(2a+1)=0\)

we get \(a=1, a=-\frac{1}{2}\)

i.e.

\(\cos(x)=1, \cos(x)=-\frac{1}{2}\)

when is this true, within the interval given?