The variables x and y satisfy the differential equation
\(\cos 2x \frac{dy}{dx} = \frac{4 \tan 2x}{\sin^2 3y}\),
where \(0 \leq x < \frac{1}{4}\pi\). It is given that \(y = 0\) when \(x = \frac{1}{6}\pi\).
Solve the differential equation to obtain the value of x when \(y = \frac{1}{6}\pi\). Give your answer correct to 3 decimal places.
Solution
First, separate the variables:
\(\sin^2 3y \, dy = 4 \sec 2x \tan 2x \, dx\).
Integrate both sides:
\(\int \sin^2 3y \, dy = \int 4 \sec 2x \tan 2x \, dx\).
The left side integrates to:
\(\frac{1}{2}y - \frac{1}{12}\sin 6y\).
The right side integrates to:
\(2 \sec 2x\).
Using the initial condition \(y = 0\) when \(x = \frac{1}{6}\pi\), find the constant of integration:
\(\frac{1}{2}(0) - \frac{1}{12}\sin(0) = 2 \sec \left(\frac{1}{3}\pi\right) - C\).
\(C = 2 \sec \left(\frac{1}{3}\pi\right)\).
Substitute back to find \(x\) when \(y = \frac{1}{6}\pi\):
\(\frac{1}{2}\left(\frac{1}{6}\pi\right) - \frac{1}{12}\sin\left(\frac{1}{2}\pi\right) = 2 \sec 2x - 2 \sec \left(\frac{1}{3}\pi\right)\).
Solve for \(x\):
\(x = 0.541\).
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