Here I think would be my answer to this question, if something is wrong, please let me know and we can figure out that~ Under this situation, we know as long as m (the angle theta) satisfies sin2m=-1, it is the solution and now 2m= 3*pi/2 + 2*k*pi, k is an integer so m=3*pi/4 + k*pi, k is an integer we now see that the terminal side of m is the bisector of the 2nd (II) and 4th (IV) quadrant, if in the 2nd quadrant, sinm=sqrt(2)/2 and cosm= - sqrt(2)/2 we have sinm-cosm=sqrt(2) if in the 4th quadrant, sinm= - sqrt(2)/2 and cosm= sqrt(2)/2, we have sinm-cosm= - sqrt(2). If x doesn't equal to 0, we divide x on both sides and we have: figure out the only situation to hold the equation true is that -1/sinm*cosm=2, that means (-1*2)/(sin2m)=2 and sin2m must be -1 to keep this true. Now it turns into a problem solving x: if x=0, it satisfies the equation and this means sinm=cosm these angles can be the bisector of the 1st (I) and 3rd(III) quadrant they can be pi/4 and 3pi/4 as you mentioned before and 2k*pi applied on them as well, under this condition, we choose 0 as the answer Let x = sinm - cosm, which is the value we try to find we have -x/(sinm*cosm)=2x Then move the terms and change it into: 1/sinm -1/cosm = 2(sinm - cosm) simplify this as: (cosm-sinm)/(sinm*cosm)=2(sinm-cosm) Then divide sinm on both sides (if sinm=0, the equation cannot be satisfied since left=0, right=0-1=-1, not equal to the left), then we have: 2cosm=1/cosm -1/sinm +2sinm Actually not just these two radians but a set of them are the solutions to the 1st equation Check this:įirst, I use m to represent theta, sqrt(2) as the square root of 2, for simplification: so we know sin(2m)=tan(m)-cos(2m) from the given facts, then we use the identities, we have left=sin(2m)=2*sinm*cosm=right=(sinm/cosm) - , If you try theta= pi/4 or negative pi/4 the first equation holds in both situations and you get choice A and C respectively. Thank you for this question, this is a great one I think.
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