= Solution
For this <first-order linear differential equation>, an <integrating factor> on an <interval> avoiding zeros of $\cos x$ is $\cos x$. The <product rule> then gives
$$
(y\cos x)'=\cos x,\qquad y\cos x=\sin x+C.
$$
The <initial condition> gives $3/\sqrt2=1/\sqrt2+C$, hence $C=\sqrt2$. Thus
$$
\boxed{y(x)=\frac{\sin x+\sqrt2}{\cos x},\qquad-\frac\pi2<x<\frac\pi2.}
$$
The displayed <interval> is the maximal one containing the specified initial <point>, since the coefficient $\tan x$ is undefined at its endpoints.
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