Sandra Allison

2021-12-30

How do you solve $\mathrm{sin}x=-\frac{1}{2}$ ?

Bernard Lacey

Explanation:
$\mathrm{sin}x$ is negative in the 3-rd and 4-th quadrants. Since
$\frac{\mathrm{sin}\pi }{6}=\frac{1}{2}$
$\mathrm{sin}x$ would be $\frac{1}{2}$ for $x=\pi +\frac{\pi }{6}$ and for $x=2\pi -\frac{\pi }{6}$
The Principal so;ution is this
$x=\frac{7\pi }{6},\frac{11\pi }{6}$

karton

Take the inverse sine of both sides of the equation to extract x from inside the sine.
$x=\mathrm{arcsin}\left(-\frac{1}{2}\right)$
The exact value of $\mathrm{arcsin}\left(-\frac{1}{2}\right)$ is $-\frac{\pi }{6}$
$x=-\frac{\pi }{6}$
The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2\pi$, to find a reference angle. Next, add this reference angle to $\pi$ to find the solution in the third quadrant.
$x=2\pi +\frac{\pi }{6}+\pi$
Simplify the expression to find the second solution.
$x=\frac{7\pi }{6}$
Find the period
$2\pi$
Add $2\pi$ to every negative angle to get positives angles.
$x=\frac{11\pi }{6}$
The period of the $\mathrm{sin}\left(x\right)$ function is $2\pi$ so values will repeat every $2\pi$ radians in both directions.
, for any integer n

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