Question

Solve the equation. (Use n as an integer constant, where x = [0, a) when n = 0.) (3 tan? x - 1)(tan? x – 3) = 0 3 tan2 x - 1 x= (smaller value for n = 0) 3 tan? x – 1 x = (larger value for n = 0) X tan? x - 3 = (smaller value for n = 0) X tan2 x – 3 x (larger value for n = 0)

Answer 1

1 6 tan2a-1) (kan 2-3) = 0. tanrx-3 =0 Eanx=3. => 3 tan2x-1=0 3 tan2 = 1 tan2x = 1/3 ton = 3 tanx = a√3 tanx = + 7 Case-I can x= t I can be written in 2 ways. tare = tanx = 1/3

x = kant to . tont Ca] God. OF 30 AG By Adding periods x = kan' (Je na nez x = tant [/]ena, nez and > x=&thin nez x = -f tha: ner --- co to a positive We have to convert means of the angle in Referance angle angle by Referance angle - here we add a a - 1 + 2 = -7/ nez > ) => x= 5x + na

Case-1 tanx = 1 Split into a cases tanx = √3 tanx s-ß. = rotanto) and x fant (-03] By Adding periods x = tant crot na nez nd x = kant (83)tno, nez - x = 7 + na, nez and Convert x= 1 tot, nez... (2) 2) into a positive angle .-1.50 -73 39-X

(2) = x= 22 +0h, nez. so we have 3 kan2x-1=0 1 x = + na, nez x = 8 + na, nor and. Ean2x3 =0 x = a +na, nez cal x= 2h+nh, nez in (3) cohen we put n=0 5) 212 711 +0. and x = 57 to X=116 x = 5516 [smaller value] [larger value]

put neo in Cw & Glm telo X+ Dato Tyd x=9 [ Smaller value x=24 C larger value]