The following function models the length $L$ of each day (in minutes) in Manila, $t$ days after the spring equinox, which is March 22.
$$ L(t) = 52 \sin\left(\frac{2\pi t}{365}\right)+728$$
What is the first time after March 22 that the day length hits 700 minutes?
what I have tried:
So I get $t = -28.5$ which is negative and days cannot be in negative
please help me in solving the problem
Update:
In the Image above sin-1 is $\sin^{-1}$
Hint:
The first 700-minute day is 216 days after March 22
On
Trigonometric aspects have been discussed in comments and answer; so, I shall propose another possible approach.
You need to solve the equation $$L(t) = 52 \sin\left(\frac{2\pi t}{365}\right)+728=700$$ and you are looking for the closest solution; so, let us write the equation as $$f(t)=52 \sin\left(\frac{2\pi t}{365}\right)+28$$ and use Newton method with $t_0=\frac{365}{2}$ which is the middle point between the maximum and the minimum of the function.
Newton iterates are given by $$t_{n+1}=t_n-\frac{f(t_n)}{f'(t_n)}$$ which in the case of the problem gives $$t_{n+1}=t_n-\frac{365 \left(52 \sin \left(\frac{2 \pi t_n}{365}\right)+28\right) \sec \left(\frac{2 \pi t_n}{365}\right)}{104 \pi }$$ Then, the following iterates are generated : $213.780$, $215.515$, $215.531$ which is the solution for six significant figures.
$52\sin \left(\frac{2\pi\cdot t}{365}\right) + 728 = 700 \to \sin \left(\frac{2\pi\cdot t}{365}\right) = \dfrac{700-728}{52} = -0.538 \to \dfrac{2\pi\cdot t}{365} = \sin^{-1} (-.538) + 2\pi = -0.57+2\pi = 5.72 \to t = \dfrac{5.72\cdot 365}{2\pi} = 332$ ( days ).
Edit:To get $216$ days, all you need to do is change $-0.57$ to $-(\pi - 0.57)$, and then add $2\pi$ to it and get $-(\pi - 0.57) + 2\pi = \pi + 0.57 = 3.71$. Thus:
$ t= \dfrac{3.71\cdot 365}{2\pi} \approx 216$ (days).