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

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Answer 1 · 2016-09-11T17:33:26

$x=npi+(-1)^npi/6$ , where $n$ is an integer.

As $2sinx-1=0$

$2sinx=1$ or

$sinx=1/2=sin(pi/6)$

Now as sine function is positive in first and second quadrant

Hence $x=pi/6$ or $x=pi-pi/6=(5pi)/6$

But as sine function has cycle of $2pi$

$x=2npi+pi/6$ or $x=2npi+(5pi)/6$

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

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