# Thread: Solve sin(x+pi/3) = -1 for 0 ≤ x ≤2pi

1. ## Solve sin(x+pi/3) = -1 for 0 ≤ x ≤2pi

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.

2. Originally Posted by jnow2
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.

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?

3. 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!

4. Originally Posted by jnow2
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!
I follow what you're trying to say. Even if you do it that way, you should still end up with $\displaystyle \tfrac{3\pi}{2}-\tfrac{\pi}{3}=\tfrac{7\pi}{6}$ as the only solution over $\displaystyle 0\leq x\leq 2\pi$.

5. Originally Posted by jnow2
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!

Can somebody please explain this method better to me? Perhaps by providing an example? I am really stuck on this.