0
$\begingroup$

Solve for $x$ in $[-\pi,\pi]$
$\cos x=\frac{-\sqrt{3}}{2}$
enter image description here
If I look up above Unit Circle,I can see that $\cos x=\frac{-\sqrt{3}}{2}$ is $\frac{5\pi}{3}$ but the right answer is $\frac{-5\pi}{6}$ or $\frac{5\pi}{6}$

Appreciate your help. thx

  • 3
    Just read: for $x=5\pi/3$, $\cos(x)=1/2$, not $-\sqrt3/2$.2012-05-01
  • 0
    ok! i got it one, but the other is when I look to $\frac{-\sqrt{3}}{2}$ it is not cos anymore, it is in $sin$ area.2012-05-01
  • 0
    The x in "cos x" is NOT the same as the x of the horizontal axis or of the ordered pair (x, y). This can be confusing. The point where angle *t*, in standard position, crosses the unit circle is $(\cos t, \sin t)$. I would recommend "Solve for *t* in $[-\pi, \pi]$"...2012-05-01
  • 0
    @TheChaz so does the $x$ mean green lines from the above graph.thx2012-05-01
  • 1
    @SbSangpi: The facts $\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin(\theta)$ are useful.2012-05-01

1 Answers 1

3

You want to find all the values of $x \in [-\pi, \pi]$ for which $\cos(x) = \frac{-\sqrt{3}}{2}$. This means that you want to find the $x$-coordinates on the unit circle for which this holds, but they have to be in the interval $[-\pi, \pi]$

Let's consider $[0, \pi]$, the upper half of the unit circle. By looking at the unit circle, we have $\cos(x) = \frac{-\sqrt{3}}{2}$ (the $x-$coordinate is $\frac{-\sqrt{3}}{2}$) at $\frac{5 \pi}{6}$.

Now, let's consider $[-\pi, 0]$, the lower half of the unit circle. $\cos(x) = \frac{-\sqrt{3}}{2}$ at $\frac{7 \pi}{6}$, but this isn't part of the interval $[-\pi, 0]$. However, since $\frac{7 \pi}{6} = \frac{7 \pi}{6} - 2\pi = -\frac{5 \pi}{6}$, and this is part of the interval $[-\pi, 0]$, $-\frac{5 \pi}{6}$ is the second value that satisfies your equation.