How were tangent terms derived in solving for eqn (2) in a second-order circuit?

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SUMMARY

The discussion focuses on the derivation of tangent terms in solving equation (2) of a second-order circuit. The author equated equation (2) to zero and extracted the common factor of 5(cos 21.749t) to simplify the problem. This step is crucial for understanding how tangent terms are derived in the context of circuit analysis. The clarification sought by the user highlights the importance of recognizing common factors in trigonometric equations.

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Electrical engineering students, circuit designers, and anyone involved in analyzing second-order circuits will benefit from this discussion.

paulmdrdo
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Homework Statement
Find the expression for ##v(t)##
Relevant Equations
SEE ATTACHED PHOTO
I was trying to follow the solution for this problem and got stuck in the last portion of the solution. I encircled the part that I did not understand. I can't figure out how the solver was able to come up with tangent terms. Please enlighten me. TIA
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The author equated eqn (2) to 0
then extracted (and divided both sides by) the common factor 5(cos 21.749t)
 

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