Method of characteristic

I have the equation

$$\partial_t u(t,x) + \sin(t) \partial_x u(t,x) +\cos(t) u(t,x) = 0.$$

with a initial condition $$u(0,x)=f(x)$$. I want to solve this using the method of characteristics. First I consider just the equation

$$\partial_t u(t,x) + \sin(t) \partial_x u(t,x)=0,$$ parametrizing by $$s$$ we get the equations

$$\frac{\mathrm d t(s)}{\mathrm d s} = 1, \quad \frac{\mathrm d x(s)}{\mathrm d s} = \sin(t(s)),$$ which has the solutions $$t(s) = s, \quad x(s) = c_1 - \cos(s).$$ with $$c_1$$ being a constant (I've determined the constant in $$t(s)$$ to be zero since we have the initial conditions at $$t=0$$). Thus the solution is $$u(t,x) = f(x + \cos(t)).$$ Now along the characteristic curves the PDE reduces to an ODE

$$\frac{\mathrm d u(s)}{\mathrm d s} + \cos(s) u(s)=0.$$

the argument for this is that $$\frac{\mathrm d u(s)}{\mathrm d s} = \frac{\mathrm d u(t(s),x(s))}{\mathrm ds} = \partial_{t(s)} u(t(s),x(s)) + \sin(t(s)) \partial_{x(s)} u(t(s),x(s)).$$

the ODE has the solution

$$u(s) = e^{-\sin(s)} c_2.$$

It's not now obvious to me where I went wrong, I assume that probrably we cannot take $$\cos(t) = \cos(s)$$ when we are along the characteristic curves. But how do I solve the equation correctly?

The function $$u(s)$$ satisfies the differential equation $$\frac{du(s)}{ds}=-u(s)\cos(t(s))=-u(s)\cos(s), \tag{1}$$ where the second equality follows from $$t(s)=s$$. The solution to $$(1)$$ is $$u(s)=u(0)\,e^{-\sin(s)}. \tag{2}$$ Now $$u(0)=u(t(0),x(0))=u(0,c_1-1)=f(c_1-1). \tag{3}$$ Solving $$t(s) = s, \qquad x(s) = c_1 - \cos(s) \tag{4}$$ for $$c_1$$ and $$s$$, we get $$s = t, \qquad c_1 = x+\cos(t). \tag{5}$$ Combining $$(2), (3)$$ and $$(5)$$, we finally obtain $$u(t,x)=f(x+\cos(t)-1)\,e^{-\sin(t)}. \tag{6}$$
$$\frac{\partial u}{\partial t}+\sin(t)\frac{\partial u}{\partial x}=-\cos(t)u$$ Characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{\sin(t)}=\frac{du}{-\cos(t)u}=ds$$ A first characteristic equation comes from solving $$\frac{dt}{1}=\frac{dx}{\sin(t)}$$ : $$x+\cos(t)=c_1$$ A second characteristic equation comes from solving $$\frac{dt}{1}=\frac{du}{-\cos(t)u}$$ : $$u\:e^{\sin(t)}=c_2$$ The general solution of the PDE on the form of implicit equation $$c_2=\Phi(c_1)$$ is :
$$u\:e^{\sin(t)}=\Phi\big(x+\cos(t)\big)$$ $$\Phi$$ is an arbitrary function (to be determined according to the initial condition). $$u(x,t)=e^{-\sin(t)}\Phi\big(x+\cos(t)\big)$$ With condition $$u(0,x)=f(x)$$ : $$u(0,x)=e^{-\sin(0)}\Phi\big(x+\cos(0)\big)=\Phi\big(x+1\big)=f(x)$$ This implies $$\Phi(X)=f(X-1)$$ .
The function $$\Phi$$ is determined. We put it into the above general solution where $$X=x+\cos(t)$$ $$u(x,t)=e^{-\sin(t)}f\big(x+\cos(t)-1\big)$$ This is the particular solution of the PDE which satisfies the initial solution.