Can the equation be solved ?

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1 Answer
Apr 13, 2018

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.)