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prove that
tan ( 2 tan^-l x)=2 tan ( tan^-1 x + tan^-1 x^3)

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.

it reflected because of the glass maube

step-by-step explanation:

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this