Question

How do you solve the following equation `sin 2x - sin x = 0` in the interval [0, 2pi]?

Answer 1

Use the identity `sin2theta = 2sinthetacostheta`.

Explanation:

`2sinxcosx - sinx = 0`

`sinx(2cosx - 1) = 0`

`sinx = 0 and cosx = 1/2`

`0, pi, pi/3 and (5pi)/3`

Hopefully this helps!

Answer 2

The soln. set `={0,pi/3,pi,5pi/3,2pi}.`

Explanation:

Given that, `sin2x-sinx=0. rArr 2sinxcosx-sinx=0. rArr sinx(2cosx-1)=0. rArr sinx=0, or, cosx=1/2`

The zeros of sin are, `kpi, k in ZZ.` Since we require zeros in `[0,2pi]`, these are `0,pi,2pi.`

Next `cosx=1/2=cos(pi/3) rArr x=2kpi+-pi/3, k in ZZ.`This is since that the general soln. of the eqn. `costheta=cosalpha` is `theta=2kpi+-alpha, k in ZZ`.

`k=0, rArr x=+pi/3 in [0,2pi],` as `-pi/3!in[0,2pi]`
`k=1 rArr x=2pi-pi/3=5pi/3,` as `2pi+pi/3 !in [0,2pi].`

Altogether, the soln. set `={0,pi/3,pi,5pi/3,2pi}.`