Combining 2 sinusoids of equal amplitude with different frequencies into 1 new "wave"
Let us assume the following signal form $$x(t)=\sin(2\pi t f(t))$$ where $f(t)$ is some "instantaneuous frequency", which should increase linearly from $w_1$ to $w_2$ over the time. 

The time should end, when the sine wave has done one period, i.e. its argument became $2\pi$. At this time, $f(t)=w_2$. From that we can calculate:

$$2\pi w_2 t_e=2\pi\\t_e=1/w_2,$$ where $t_e$ is the end time of the simulation. Hence, we have $$f(w)=w_1+\frac{w_2-w_1}{t_e}t=w_1+(w_2-w_1)w_2t.$$ We can plot this function:

w1 = 5
w2 = 10
T = 1/w2

Fs = 100000

t = np.arange(0, T, 1/Fs)

f = np.linspace(w1, w2, len(t))
plt.plot(t, np.sin(2*np.pi*f*t))
plt.title('Graph')
plt.grid(True)




Now, regarding the question, where the zero is. The sine goes to zero, when its argument becomes $\pi$. Hence, we have

$$2\pi (w_1+(w_2-w_1)w_2t)t=\pi,$$

which we can solve by (its a quadratic equation, so nothing special, but I use mathematica for convenience):

$$\text{Solve}\left[t (t \text{w2} (\text{w2}-\text{w1})+\text{w1})=\frac{1}{2},t\right]$$

$$\left\{\left\{t\to \frac{\sqrt{\text{w1}^2-2 \text{w1} \text{w2}+2 \text{w2}^2}-\text{w1}}{2 \left(\text{w2}^2-\text{w1} \text{w2}\right)}\right\},\left\{t\to \frac{\sqrt{\text{w1}^2-2 \text{w1} \text{w2}+2 \text{w2}^2}+\text{w1}}{2 \left(\text{w1} \text{w2}-\text{w2}^2\right)}\right\}\right\}$$

There are two solutions, because for negative t, we can also end up with a solution (both $f(t)$ and $t$ would be negative to yield 1/2 as the argument).

To answer the comment from @Jason R: This indeed can also be seen as a frequency-modulation, given by

$$x(t)=\sin\left(2\pi\int_0^t\phi(\tau)d\tau\right)$$

when setting

$$\phi(\tau)=w_1+2(w_2-w_1)w_2\tau.$$

In this case we have $tf(t)=\int_0^t\phi(\tau)d\tau$.