Solve for $x$ in $[-\pi,\pi]$
$\cos x=\frac{-\sqrt{3}}{2}$
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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
Solve for $x$ in $[-\pi,\pi]$
$\cos x=\frac{-\sqrt{3}}{2}$
![]()
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
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.