Solve for x
3cos(x)+2sin2(x)=0
3cos(x)+2sin2(x)=03cos(x)=−2sin2(x)|:(−2)−1.5cos(x)=sin2(x)
Substitute:
t=tan(x2)cos(x)=1−t21+t2sin(x)=2t1+t2
−1.5cos(x)=sin2(x)|cos(x)=1−t21+t2, sin(x)=2t1+t2−1.5(1−t2)(1+t2)=4t2(1+t2)2−1.5(1−t2)=4t21+t2−1.5(1−t2)(1+t2)=4t2−1.5(1−t4)=4t2−1.5+1.5t4=4t21.5t4−4t2−1.5=0t2=4±√16−4∗(1.5)∗(−1.5)2∗1.5t2=4±√16+93t2=4±√253t2=4±53t21=4+53t21=93t21=3t22=4−53t22=−13
x= ?x= ?tan(x2)=t1tan(x2)=t2tan(x2)=√3tan(x2)=√−13(the root is imaginary !)
tan(x2)=√3tan(x2)=√3tan(x2)=+√3tan(x2)=−√3x2=arctan(√3)+180∘kk∈Zx2=arctan(−√3)+180∘kk∈Zx2=60∘+180∘kk∈Zx2=−60∘+180∘kk∈Zx=120∘+360∘kk∈Zx=−120∘+360∘kk∈Zx=360∘−120∘+360∘kk∈Zx=240∘+360∘kk∈Z