Answer
Either $t=0.142$ or $t=0.785$
Work Step by Step
$$s(t)=\sin t+2\cos t$$
For $s(t)=\frac{3\sqrt2}{2}$, we would have the equation:
$$\sin t+2\cos t=\frac{3\sqrt2}{2}$$
- We do not have any identities that can relate $\sin t$ and $\cos t$, but we have this identity: $\sin^2 t+\cos^2t=1$, which might be useful.
So the strategy now is to square both sides. Yet before doing so, we rearrange the equation:
$$\sin t=\frac{3\sqrt2}{2}-2\cos t$$
Now square both sides:
$$\sin^2t=\frac{(3\sqrt2)^2}{4}-2\times\frac{3\sqrt2}{2}\times2\cos t+4\cos^2t$$
$$\sin^2t=\frac{9}{2}-6\sqrt2\cos t+4\cos^2t$$
Now apply $\sin^2t=1-\cos^2t$:
$$1-\cos^2t=\frac{9}{2}-6\sqrt2\cos t+4\cos^2t$$
$$5\cos^2t-6\sqrt2\cos t+\frac{7}{2}=0$$
We can treat the question as a quadratic formula, which $a=5, b=-6\sqrt2, c=\frac{7}{2}$
- Find $\Delta$: $$\Delta=b^2-4ac$$
$$\Delta=(-6\sqrt2)^2-4\times5\times\frac{7}{2}$$
$$\Delta=36\times2-2\times5\times7$$
$$\Delta=72-70$$
$$\Delta=2$$
- Find $\cos t$: $$\cos t=\frac{-b\pm\sqrt\Delta}{2a}=\frac{6\sqrt2\pm\sqrt{2}}{10}$$
1) For $\cos t=\frac{6\sqrt2+\sqrt{2}}{10}=\frac{7\sqrt2}{10}\approx0.99$
$$t=\cos^{-1}0.99\approx0.142$$
- Try back to $s(t)$:
$$\sin0.142+2\cos0.142\approx2.121=\frac{3\sqrt2}{2}$$
Therefore, $t=0.142$ would be accepted.
2) For $\cos t=\frac{6\sqrt2-\sqrt{2}}{10}=\frac{5\sqrt2}{10}=\frac{\sqrt2}{2}$
$$t=\cos^{-1}\frac{\sqrt2}{2}=\frac{\pi}{4}$$
- Try back to $s(t)$:
$$\sin\frac{\pi}{4}+2\cos\frac{\pi}{4}=\frac{\sqrt2}{2}+2\times\frac{\sqrt2}{2}=\frac{3\sqrt2}{2}$$
Therefore, $t=\frac{\pi}{4}\approx0.785$ would also be accepted.
Yet the question asks for only one value of $t$, so you can either choose $t=0.142$ or $t=0.785$
