Answer:
The answers are 0, 1 and −2.
Step-by-step explanation:
Let α=arctan(2tan2x) and β=arcsin(3sin2x5+4cos2x).
sinβ2tanβ21+tan2β2tanβ21+tan2β23tanxtan2β2−(9+tan2x)tanβ2+3tanx(3tanβ2−tanx)(tanβ2tanx−3)tanβ2=3sin2x5+4cos2x=3(2tanx1+tan2x)5+4(1−tan2x1+tan2x)=3tanx9+tan2x=0=0=13tanxor3tanx
Note that x=α−12β.
tanx=tan(α−12β)=tanα−tanβ21+tanαtanβ2=2tan2x−13tanx1+2tan2x(13tanx)or2tan2x−3tanx1+2tan2x(3tanx)=tanx(6tanx−1)3+2tan3xor2tan3x−3tanx(1+6tanx)
So we have tanx=0, tan3x−3tanx+2=0 or 4tan3x+tan2x+3=0.
Solving, we have tanx=0, 1, −1 or −2.
Note that −1 should be rejected.
tanx=−1 is corresponding to tanβ2=3tanx. So tanβ2=−3, which is impossible as β∈[−π2,π2].
The answers are 0, 1 and −2.
