Question

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

Answer 1

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

Explanation:

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}`