Question

HELP NEED PRECALC HELP WILL GIVE BRAINLIEST PLEASE HELP

Answer 1

From your earlier questions, we found

$2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)$

so the wave has amplitude √29. The weight's maximum negative position from equilibrium is then -√29, so you are solving for t in the given interval for which

$\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)=-\dfrac{\sqrt{29}}2$

Divide both sides by √29:

$\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)=-\dfrac12$

Take the inverse sine of both sides, noting that we get two possible solution sets because we have

$\sin\left(\dfrac{7\pi}6\right)=\sin\left(\dfrac{11\pi}6\right)=-\dfrac12$

and the sine wave has period 2π, so $\sin x=\sin(x+2\pi)=\sin(x+4\pi)=\cdots$.

$\implies 4\pi t+\tan^{-1}\left(\dfrac52\right)=\dfrac{7\pi}6+2n\pi$

OR

$\implies 4\pi t+\tan^{-1}\left(\dfrac52\right)=\dfrac{11\pi}6+2n\pi$

where n is any integer.

Now solve for t :

$t=\dfrac{\frac{7\pi}6+2n\pi-\tan^{-1}\left(\frac52\right)}{4\pi}$

OR

$t=\dfrac{\frac{11\pi}6+2n\pi-\tan^{-1}\left(\frac52\right)}{4\pi}$

We get solutions between 0 and 0.5 when n = 0 of t ≈ 0.196946 and t ≈ 0.363613.