Solve this exact differential equation $$[2\sin(2x)\sin(hy)] \, dx-[\cos(2x)\cos(hy)] \, dy = 0$$
This exact differential equation should give a solution something like $\sin(hy)\cos(2x)=\sin(h+1)$ when an initial value of $y(0)=1$ is applied.

$$[2\sin(2x)\sin(hy)]\,dx-[\cos(2x)\cos(hy)]\,dy = 0$$
Since this differential equation is exact we can figure out what is $h$
$$\frac{\partial (2\sin(2x)\sin(hy))}{\partial y}=2h\sin(2x)\cos(hy)$$
$$\frac{\partial (-\cos(2x)\cos(hy))}{\partial x}=2\sin(2x)\cos(hy)$$
$$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$$
$$2h=2, h=1$$
Notice that the function is separable,
$$\frac{2\sin(2x)}{\cos(2x)}\,dx-\frac{\cos(hy)}{\sin(hy)}\,dy=0$$
$$\int\frac{2\sin(2x)}{\cos(2x)}\,dx-\int \frac{\cos(hy)}{\sin(hy)}\,dy=\int 0 $$
$$-\ln|\cos2x|-\frac{1}{h}\ln|\sin(hy)|=c$$
Applying $x=0,y=1$
$$-\ln|\cos0|-\frac{1}{h}\ln|\sin(h)|=c$$
$$c=-\frac{1}{h}\ln\sin(h)$$
$$-\ln|\cos2x|-\frac{1}{h}\ln|\sin(hy)|=-\frac{1}{h}\ln\sin(h)$$
Can you take it from here?
Addendum:
Recall that
$$\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)| $$
$$\int \cos x \,dx=\sin x$$
$$\int \sin x \,dx=-\cos x$$

$$\frac{dy}{dx}=2\tan(2x)\tan(hy)\qquad\text{is a separable ODE.}$$
$$\int \cot(hy)dy=2\int\tan(2x)dx$$
$$\frac{1}{h}\ln|\sin(hy)|=-\ln(\cos(2x)|+c$$
$$\sin(hy)(\cos(2x))^{h}=e^{hc}=C$$
Condition $y(0)=1 \quad\to\quad \sin(h)(\cos(0))^{h}=\sin(h)=C$
$$(\sin(hy))(\cos(2x))^{h}=\sin(h)$$
$$y=\frac{1}{h}\arcsin\left(\frac{\sin(h)}{(\cos(2x))^{h}} \right)$$