I need to find where
20cos(5t) + 4sin(5t) = 0 for the first and second time after the initial t = 0
I get to
arctan (-5) = 5t
I can take arctan of -5 to get -1.3... then divide by 5, but that doesn't help me.
I need help quickly please.
I need to find where
20cos(5t) + 4sin(5t) = 0 for the first and second time after the initial t = 0
I get to
arctan (-5) = 5t
I can take arctan of -5 to get -1.3... then divide by 5, but that doesn't help me.
I need help quickly please.
ok first rewrite this as a single trig function with a phase offset.
$20\cos(5t)+4\sin(5t)=A\cos(5t+\phi)$
$A=\sqrt{20^2+4^2}=\sqrt{416}$
$\phi = \arctan\left(\dfrac{4}{20}\right)\approx 0.197~ rad$
Now the cosine will be 0 for all odd multiples of $\dfrac{\pi}{2}$ so we look for
$5t+\phi=\dfrac{2k+1}{2}\pi$
$t=\dfrac{1}{5}\left(\dfrac{2k+1}{2}\pi-\phi\right)$
you can probably take it from here.
Your first way of attacking it would have worked but you goofed somewhere.
$20\cos(5t)+4\sin(5t)=0$
$20\cos(5t)=-4\sin(5t)$
$-5=\tan(5t)$
ah I see what happened. You just took the arctan here ignoring the fact that the tangent function is periodic. If you do this
$-5 = \tan(5t+k\pi)$ and solve for $k=0,1$ you'll arrive at the same answers.