Ok. So this question has taken me almost 2 hours. There's no solution in my book and I honestly cannot figure out how to do.
The only answer I can come up with is that there is no solutions. Anyone care to please help?
Ok. So this question has taken me almost 2 hours. There's no solution in my book and I honestly cannot figure out how to do.
The only answer I can come up with is that there is no solutions. Anyone care to please help?
There are solutions...
Note that $\displaystyle \sin\left(x+\tfrac{\pi}{3}\right)=-1\implies x+\tfrac{\pi}{3}=\arcsin(-1)$.
Now, $\displaystyle \arcsin(-1)=\tfrac{3\pi}{2}+2k\pi,\,k\in\mathbb{Z}$.
Therefore, $\displaystyle x=\tfrac{3\pi}{2}-\tfrac{\pi}{3}+2k\pi=\tfrac{7\pi}{6}+2k\pi,\,k\in\ mathbb{Z}$.
The only solution that falls in the interval $\displaystyle 0\leq x\leq 2\pi$ is $\displaystyle \tfrac{7\pi}{6}$ (Do you see why?).
Does this make sense?
not exactly. thanks so much for the speedy reply but i'm still extremely unsure of what you did as we have been taught differently. Infact i'm not even sure if we've been taught arcsin yet.
We have been taught so far to draw up a circle and mark where -1 lies by looking up a radians chart (it sits on 270 degrees)
We then use rotations to make plots, for example, the rotation from 0-270 degrees is 3pi/2 so that becomes the first point. Then we continue in the positive direction and then we go in the negative direction to make plots.
We then subtract pi/3 from each plot and select which ones lie within the above mentioned greater than/ less than bracket. Sorry, that's probably extremely confusing!