How do you solve $2sinx cosx + cosx = 0$ from 0 to 2pi?

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Answer 1 · 2016-02-24T10:45:23

Solution set is ${pi/2, (7pi)/6, (3pi)/2, (11pi)/6}$

In $2sinxcosx+cosx=0$ , taking $cosx$ common we get

$cosx(2sinx+1)=0$

Hence,

either $cosx=0$ whose solution in domain $[0, 2pi]$ is ${pi/2, (3pi)/2}$

or $2sinx+1=0$ i.e. $sinx=-1/2$ whose solution in domain $[0, 2pi]$ is ${(7pi)/6, (11pi)/6}$

Hence solution set is ${pi/2, (7pi)/6, (3pi)/2, (11pi)/6}$

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