Question

Solve using double angles?. Sin2x-sinx=0

Answer 1

`x = pm 2 pi n, frac(pi)(3) pm 2 pi n`; where `n in ZZ`.

Explanation:

We have: `sin(2 x) - sin(x) = 0`

Using the double-angle identity for `sin(x)`, we get:

`Rightarrow 2 sin(x) cos(x) - sin(x) = 0`

`Rightarrow sin(x) (2 cos(x) - 1) = 0`

`Rightarrow sin(x) = 0`

`Rightarrow x = 0 pm 2 pi n`; `n in ZZ`

`Rightarrow x = pm 2 pi n`; `n in ZZ`

or

`Rightarrow cos(x) = frac(1)(2)`

`Rightarrow x = frac(pi)(3) pm 2 pi n`; `n in ZZ`

Therefore, the general solutions to the equation are `x = pm 2 pi n` or `x = frac(pi)(3) pm 2 pi n`, where `n in ZZ`.

Answer 2

`x=kpi" or "x=pi/3+2kpi`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)sin2x=2sinxcosx`

`2sinxcosx-sinx=0`

`sinx(2cosx-1)=0`

`"equate each factor to zero and solve for "x`

`sinx=0rArrx=kpito(k in ZZ)`

`cosx=1/2rArrx=pi/3+2kpito(k inZZ)`