MHB Solve for Theta: Equilibrium of a Particle

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The discussion focuses on solving the equation for theta related to the equilibrium of a particle. The equation presented is complex, involving trigonometric functions and requires careful manipulation. Participants suggest verifying the equation's correctness and recommend using trigonometric identities to simplify the problem, such as substituting tan(theta) with sin(theta)/cos(theta). There is also a suggestion to square the equation to facilitate solving for cos(theta). The conversation emphasizes the importance of clarity in the steps already attempted to aid further assistance.
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solve for theta

$\frac{\tan(\theta)(\sqrt{5-4\cos(\theta)}-1)}{\sqrt{5-4\cos(\theta)}}=\frac{10}{60}$

I have already tried my best solving this eqn but still couldn't get it. FYI getting that equation already took me a lot of work. Now I'am on the last piece of the problem I am solving which is to solve for theta. So please kindly tell me how to go about it. Thanks! By the way the problem is about equlibrium of a particle. Thanks!
 
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Well, you say that it took a lot of time for you to get this equation. So, I request you to cross check if the equation is correct.
 
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I think puttiing $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ shall comlpicate. I would square it and put $\tan^2\theta= \dfrac{ 1}{\cos ^2 \theta} -1$ then solve for $\cos \theta$

By the way could you mention what you have tried
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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