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

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Answer 1 · 2016-08-31T23:30:40

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 · 2016-09-04T08:08:31

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

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)#