I need to find the number of m values such that this equation has solutions.
cos4(x)+4sin2(x)+sin4(x)+4cos2(x)=m
My approach:
After I replaced sin2(x)=1−cos2(x), I noted cos2(x)=t;tfrom[0,1], right ?
I did some calculations and I got the equation in t: ∣t−2∣+∣t+1∣=m
Now I solved this equation in right way?
Case 1. t from (-infinity, -1) I got t=(1-m)/2 But T is also from [0,1] so here I have no m values such that this equation has solutions, right?
Case 2. t from (-1,2) I got t=0 so m=3 , this case is right, I have t=0 which is from [0,1] and (-1,2), right? So here I have one value for m.
Case 3. t from (2,infinity) but t is also from [0,1] so this case is similar with case 1, right ?So I have no m values.
In conclusion, the number of m values such that this equation has solutions is 1.
Is my approach correct?
cos4(x)+4sin2(x)+sin4(x)+4cos2(x)=m
My approach:
After I replaced sin2(x)=1−cos2(x), I noted cos2(x)=t;tfrom[0,1], right ?
I did some calculations and I got the equation in t: ∣t−2∣+∣t+1∣=m
Now I solved this equation in right way?
Case 1. t from (-infinity, -1) I got t=(1-m)/2 But T is also from [0,1] so here I have no m values such that this equation has solutions, right?
Case 2. t from (-1,2) I got t=0 so m=3 , this case is right, I have t=0 which is from [0,1] and (-1,2), right? So here I have one value for m.
Case 3. t from (2,infinity) but t is also from [0,1] so this case is similar with case 1, right ?So I have no m values.
In conclusion, the number of m values such that this equation has solutions is 1.
Is my approach correct?