Solve 3 - sinx = 3 + cos2x for 0° < x < 360° ?
1 Answer
Apr 27, 2018
Use
#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
#x = 180˚ + 30˚ and 360˚ - 30˚ and 90˚#
#x = 210˚, 330˚, 90˚#
We can confirm graphically. My apologies for it being in radians!
Hopefully this helps!