How do you solve sin⁡x=−12 ?

Question

Bernard Lacey · Accepted Answer

Explanation: sin⁡x is negative in the 3-rd and 4-th quadrants. Since sin⁡π6=12 sin⁡x would be 12 for x=π+π6 and for x=2π−π6 The Principal so;ution is this x=7π6,11π6

karton · Accepted Answer

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

user_27qwe · Accepted Answer

Thanks. It&#039;s helped me.

How do you solve \sin x=-\frac12 ?

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