Solving for 't': Tips and Tricks for Navigating Tricky Equations

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The discussion focuses on solving the equation 1/2 + pi/4 = (9.8/4)cos(2t) + sin(2t) for 't'. Participants express frustration with their attempts, noting that trigonometric equations often require iterative methods for solutions. A suggested approach involves rewriting the right-hand side in the form A*cos(2t + ω), utilizing trigonometric addition formulas to derive numerical values for amplitude (A) and phase (ω). This method aims to simplify the equation for easier resolution. Overall, the conversation highlights common challenges and strategies in tackling complex trigonometric equations.
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Homework Statement



Solve for 't'
1/2+pi/4 = (9.8/4)cos2t + sin(2t)

The Attempt at a Solution



I don't know what's wrong with my brain today but every attempt came up empty :P
 
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Absent some Jedi mind tricks, most trig equations require an iterative method of solution.
 
SteamKing said:
Absent some Jedi mind tricks, most trig equations require an iterative method of solution.

Nice to know the brain is still functioning properly, it's been a long day :P
 
mesa said:

Homework Statement



Solve for 't'
1/2+pi/4 = (9.8/4)cos2t + sin(2t)

The Attempt at a Solution



I don't know what's wrong with my brain today but every attempt came up empty :P

Re-write the right-hand-side in the form A*cos(2t + ω), where A is the amplitude and ω is the 'phase'. Use the trigonometric addition formulas to (eventually) get numerical values for A and ω.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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