Question

Can the equation be solved ?

Answer 1

The equation does have a solution, with `a=b≠0, theta=kpi, k in ZZ`.

Explanation:

First of all, note that `sec^2(theta)=1/cos^2(theta)≥1` for all `theta in RR`.

Then, consider the right-hand side. For the equation to have a solution, we must have

`(4ab)/(a+b)^2>=1`

`4ab>=(a+b)^2=a^2+2ab+b^2`
{since `(a+b)^2≥0` for all real `a,b`}

`0≥a^2-2ab+b^2`

`0≥(a-b)^2`

The only solution is when `a=b`.

Now, substitute `a=b` into the original equation:
`sec^2(theta)=(4a^2)/(2a)^2=1`

`1/cos^2(theta)=1`

`cos(theta)=±1`

`theta=kpi, k in ZZ`

Thus, the equation does have a solution, with `a=b≠0, theta=kpi, k in ZZ`.

(If `a=b=0`, then there would be a division-by-zero in the original equation.)