If sin⁡A=sin2⁡B and 2cos2⁡A=3cos2⁡B then the triangle ABC is

sin⁡A=sin2⁡B and 2cos2⁡A=3cos2⁡B⇒ 2−2sin2⁡A=3−3sin2⁡B⇒ 2sin2⁡A−3sin⁡A+1=0⇒ (2sin⁡A−1)(sin⁡A−1)=0⇒ A=30∘ or A=90∘ If  A=30∘⇒B=45∘⇒C=105∘ If A=90∘⇒B=90∘, which is not possible