2cos(3t-30deg)=-(square root of 2)
[0,360)
Hello,
If $\displaystyle t \in [0,360]$ then $\displaystyle 3t \in [0,1080]$.
Hence $\displaystyle 3t-30=X \in [-30,1050]$.
The problem is equivalent to finding X such as $\displaystyle 2 \cos(X)=-\sqrt{2}$
$\displaystyle \cos(X)=-\frac{\sqrt{2}}{2}$
We know that if $\displaystyle \cos(X)=\cos(a)$ then the solutions are $\displaystyle \left\{X=a+360k; X=-a+360k \ / \ k \in \mathbb{Z} \right\}$
We know that $\displaystyle \cos(315)=\frac{-\sqrt{2}}{2}$
So find all solutions in $\displaystyle [-30,1050]$ according to what i've written just before
Then replace for t in X