Solve 3 - sinx = 3 + cos2x for 0° < x < 360° ?

1 Answer
Apr 27, 2018

Use #cos(2x) = 1 - 2sin^2x#.

#3 - sinx = 3 + 1 - 2sin^2x#

#2sin^2x - sinx - 1 = 0#

#2sin^2x- 2sinx + sinx - 1 = 0#

#2sinx(sinx - 1) + (sinx- 1) = 0#

#(2sinx+ 1)(sinx- 1) =0 #

#sinx = -1/2 or 1#

Sine is negative in quadrants #III# and #IV#. It will have a value of #1/2# when the reference angle is #30˚#. Thus

#x = 180˚ + 30˚ and 360˚ - 30˚ and 90˚#

#x = 210˚, 330˚, 90˚#

We can confirm graphically. My apologies for it being in radians!

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Hopefully this helps!