Solve the equation $6 {sec}^{2} x + 3 {tan}^{2} x - 9 = 0$ $6sec^2x+3tan^2x-9=0$ in the interval $[0, 2 π]$ $[0,2pi]$ ?

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Answer 1 · 2017-03-23T17:17:46

$x = {\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}}$ $x={pi/6,(5pi)/6,(7pi)/6,(11pi)/6}$

In $6 {sec}^{2} x + 3 {tan}^{2} x - 9 = 0$ $6sec^2x+3tan^2x-9=0$ putting ${sec}^{2} x = 1 + {tan}^{2} x$ $sec^2x=1+tan^2x$ , we get

$6 (1 + {tan}^{2} x) + 3 {tan}^{2} x - 9 = 0$ $6(1+tan^2x)+3tan^2x-9=0$

or $9 {tan}^{2} x - 3 = 0$ $9tan^2x-3=0$

or $3 {tan}^{2} x - 1 = 0$ $3tan^2x-1=0$

or $(\sqrt{3} tan x + 1) (\sqrt{3} tan x - 1) = 0$ $(sqrt3tanx+1)(sqrt3tanx-1)=0$

i.e. $tan x = - \frac{1}{\sqrt{3}}$ $tanx=-1/sqrt3$ or $\frac{1}{\sqrt{3}}$ $1/sqrt3$

and in the interval $[0, 2 π]$ $[0,2pi]$

$x = {\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}}$ $x={pi/6,(5pi)/6,(7pi)/6,(11pi)/6}$

Solve the equation 6sec^2x+3tan^2x-9=06sec2x+3tan2x−9=06sec^2x+3tan^2x-9=0 in the interval [0,2pi][0,2π][0,2pi]?