Question

How do you solve `sin2xcosx-sinx=0` in the interval [0,360]?

Answer 1

Apply the identity `sin2x = 2sinxcosx`.

`2sinxcosxcosx - sinx = 0`

`2sinxcos^2x - sinx = 0`

`sinx(2cos^2x - 1) = 0`

`sinx = 0 and cosx = +-1/sqrt(2)`

`x = 0˚, 180˚, 45˚, 135˚, 225˚, 315˚`

Hopefully this helps!

Answer 2

`x = 0^(circ), 45^(circ), 135^(circ), 180^(circ), 225^(circ), 315^(circ)`

Explanation:

We have: `sin(2x) cos(x) - sin(x) = 0`; `[0^(circ),360^(circ)]`

Let's begin by applying the double-angle identity for `sin(x)`:

`=> 2 sin(x) cos(x) cdot cos(x) - sin(x) = 0`

`=> 2 sin(x) cos^(2)(x) - sin(x) = 0`

`=> sin(x)(2cos^(2)(x) - 1) = 0`

`=> sin(x) = 0 => x = 0^(circ)`

or

`=> cos(x) = pm (sqrt(2)) / (2) => x = 45^(circ), 135^(circ), 180^(circ), 225^(circ), 315^(circ)`

`=> x = 0^(circ), 45^(circ), 135^(circ), 180^(circ), 225^(circ), 315^(circ)`