Question

How do you solve `cosx+2cosxsinx=0` on the interval `[0, 2pi)`?

Answer 1

`x={pi/2,(7pi)/6,(3pi)/2,(11pi)/6}`

Explanation:

`cosx+2cosxsinx=0` can be written as

`cosx(1+2sinx)=0`

Hence either `cosx=0` i.e. `x=pi/2` or `(3pi)/2`

or `1+2sinx=0` i.e. `sinx=-1/2` i.e. `x=(7pi)/6` or `(11pi)/6`

and hence `x={pi/2,(7pi)/6,(3pi)/2,(11pi)/6}`