Question

How do you solve `sin x - 2sin x cos x = 0` over the interval 0 to 2pi?

Answer 1

`x={0,pi/3.pi,(5pi)/3,2pi}`

Explanation:

`sinx-2sinxcosx=0`

`hArrsinx(1-2cosx)=0`

Hence either `sinx=0` or `1-2cosx=0` i.e. `cosx=1/2`

Hence, `x={0,pi/3.pi,(5pi)/3,2pi}`