Question

6 cos 2t+9 cost= - 3

Answer 1

Explanation:

`6\cos \left(2t\right)+9\cos \left(t\right)=-3\\6\cos \left(2t\right)+9\cos \left(t\right)+3=0\\3+\left(-1+2\cos ^2\left(t\right)\right)\cdot \:6+9\cos \left(t\right)=0\\\\-3+12\cos ^2\left(t\right)+9\cos \left(t\right)=0\\\\-3+12u^2+9u=0\\\cos \left(t\right)=(1)/(4),\:\cos \left(t\right)=-1\\t=\arccos \left((1)/(4)\right)+2\pi n,\:t=2\pi -\arccos \left((1)/(4)\right)+2\pi n,\:t=\pi +2\pi n\\t=1.31811\dots +2\pi n,\:t=2\pi -1.31811\dots +2\pi n,\:t=\pi +2\pi n`