How do you solve $2 sin x - 1 = 0$ $2sinx-1=0$ ?

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Answer 1 · 2016-09-11T17:33:26

$x = n π + {(- 1)}^{n} \frac{π}{6}$ $x=npi+(-1)^npi/6$ , where $n$ $n$ is an integer.

As $2 sin x - 1 = 0$ $2sinx-1=0$

$2 sin x = 1$ $2sinx=1$ or

$sin x = \frac{1}{2} = sin (\frac{π}{6})$ $sinx=1/2=sin(pi/6)$

Now as sine function is positive in first and second quadrant

Hence $x = \frac{π}{6}$ $x=pi/6$ or $x = π - \frac{π}{6} = \frac{5 π}{6}$ $x=pi-pi/6=(5pi)/6$

But as sine function has cycle of $2 π$ $2pi$

$x = 2 n π + \frac{π}{6}$ $x=2npi+pi/6$ or $x = 2 n π + \frac{5 π}{6}$ $x=2npi+(5pi)/6$

We can write it in a single equation as $x = n π + {(- 1)}^{n} \frac{π}{6}$ $x=npi+(-1)^npi/6$ , where $n$ $n$ is an integer.

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