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The general solution of the equation 2sin x+1=0 is?

1 Answer

Joe B. · Hriman

Mar 8, 2018

In radians: $x = - \frac{π}{6} + 2 π n$ $x=-pi/6+ 2pin$ or $x = - \frac{5 π}{6} + 2 π n$ $x=-(5pi)/6 +2pin$ , $n \in Z$ $n∈Z$

Explanation:

$2 sin x + 1 = 0$ $2sinx + 1= 0$

$2 sin x = - 1$ $2sinx = -1$

$sin x = - \frac{1}{2}$ $sinx = -1/2$

$x = {sin}^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ $x =sin^-1(-1/2) = -pi/6$ or $- \frac{5 π}{6}$ $-(5pi)/6$

General solution: $x = - \frac{π}{6} + 2 π n$ $x=-pi/6+ 2pin$ or $x = - \frac{5 π}{6} + 2 π n$ $x=-(5pi)/6 +2pin$ , $n \in Z$ $n∈Z$

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