Find particular solution for DE, not getting the correct answer

Question

$\sin(x)\cos(2y)\,dx + \cos(x)\sin(2y)\,dy = 0$ with initial conditions $y(0)=\frac{\pi}{2}$

I got that the general family of solutions was $\log|\cos x|=-\frac{1}{2}\log|\cos 2y|+C$

When I put in the initial conditions, I got $C=0$

When I substituted in the $C$ I got $1=\cos^2(x)\cos(2y)$.

The correct answer is $-1=\cos^2(x)\cos(2y)$.

Answer 1

$$\log|\cos x|=-\frac{1}{2}\log|\cos 2y|+C\qquad \text{OK}$$ The condition $y(0)=\frac{\pi}{2}$ implies $C=0$ $$2\log|\cos x|+\log|\cos 2y|=0$$ $$\cos^2(x)\:|\cos(2y)|=1$$ $$\cos^2(x)\cos(2y)=\pm 1$$ The condition $y(0)=\frac{\pi}{2}\quad \to \quad \cos(2y)=-1\quad$ implies : $$\cos^2(x)\cos(2y)=-1$$

Answer 2

The DE can be rearranged into $$\tan(x)dx=-\tan(2y)dy$$ We can find that the $\tan x$ integral is: $$\int\tan xdx=\ln C-\ln|\cos x|$$ So therefore: $$\ln C-\ln|\cos x|=\frac{1}{2}\ln|\cos(2y)|$$ Or $$\ln C=\ln\left|\cos x\sqrt{\cos(2y)}\right|$$ And so we obtain the equation: $$\left|\cos x\sqrt{\cos(2y)}\right|=C$$ If we sub in $\{y=\frac{1}{2}\pi,x=0\}$, then we see that: $$|-1|=C\rightarrow C=1$$ So the full solution is: $$y=\frac{1}{2}\cos^{-1}\left(\frac{1}{\cos^2 x}\right)$$