Answer
$$t=\frac{\pi}{3}\approx1.407$$
Work Step by Step
$$s(t)=\sin t+2\cos t$$
For $s(t)=\frac{2+\sqrt3}{2}$, we would have the equation:
$$\sin t+2\cos t=\frac{2+\sqrt3}{2}$$
- 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:
$$\sin t=\frac{2+\sqrt3}{2}-2\cos t$$
Now square both sides:
$$\sin^2t=\frac{(2+\sqrt3)^2}{4}-2\times\frac{2+\sqrt3}{2}\times2\cos t+4\cos^2t$$
$$\sin^2t=\frac{7+4\sqrt3}{4}-2(2+\sqrt3)\cos t+4\cos^2t$$
Now apply $\sin^2t=1-\cos^2t$:
$$1-\cos^2t=\frac{7+4\sqrt3}{4}-2(2+\sqrt3)\cos t+4\cos^2t$$
$$5\cos^2t-2(2+\sqrt3)\cos t+\frac{3+4\sqrt3}{4}=0$$
We can treat the question as a quadratic formula, which $a=5, b=-2(2+\sqrt3), c=\frac{3+4\sqrt3}{4}$
- Find $\Delta$: $$\Delta=b^2-4ac$$
$$\Delta=4(2+\sqrt3)^2-4\times5\times\frac{3+4\sqrt3}{4}$$
$$\Delta=4(7+4\sqrt3)-5(3+4\sqrt3)$$
$$\Delta=28+16\sqrt3-15-20\sqrt3$$
$$\Delta=13-4\sqrt3$$
- Find $\cos t$: $$\cos t=\frac{-b\pm\sqrt\Delta}{2a}=\frac{2(2+\sqrt3)\pm\sqrt{13-4\sqrt3}}{10}$$
1) For $\cos t=\frac{2(2+\sqrt3)+\sqrt{13-4\sqrt3}}{10}\approx0.993$
$$t=\cos^{-1}0.993\approx0.12$$
- Try back to $s(t)$:
$$\sin0.12+2\cos0.12\approx2.105\ne\frac{2+\sqrt3}{2}$$
Therefore, $t=0.12$ would be eliminated.
2) For $\cos t=\frac{2(2+\sqrt3)-\sqrt{13-4\sqrt3}}{10}=\frac{1}{2}$
$$t=\cos^{-1}\frac{1}{2}=\frac{\pi}{3}$$
- Try back to $s(t)$:
$$\sin\frac{\pi}{3}+2\cos\frac{\pi}{3}=\frac{\sqrt3}{2}+2\times\frac{1}{2}=\frac{\sqrt3}{2}+1=\frac{2+\sqrt3}{2}$$
Therefore, $t=\frac{\pi}{3}$ would be accepted.
In conclusion, $$t=\frac{\pi}{3}\approx1.407$$
