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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

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3  
Just read: for $x=5\pi/3$, $\cos(x)=1/2$, not $-\sqrt3/2$. –  Did May 1 '12 at 12:46
    
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. –  Sb Sangpi May 1 '12 at 12:55
    
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 $[-\pi, \pi]$"... –  The Chaz 2.0 May 1 '12 at 13:04
    
@TheChaz so does the $x$ mean green lines from the above graph.thx –  Sb Sangpi May 1 '12 at 13:13
1  
@SbSangpi: The facts $\cos(-\theta)=\cos(\theta)$ and $\sin(-\theta)=-\sin(\theta)$ are useful. –  André Nicolas May 1 '12 at 14:48

1 Answer 1

up vote 4 down vote accepted

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.

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