Question

For 0 ≤ x ≤ 2pi, solve the equation:tanx = 4sec^2x-4

Answer 1

Given the equation;

$\tan x=4\sec ^2x-4$

We start by moving all terms to the left side of the equation;

$\tan x-4\sec ^2x+4=0$

Now we re-write this using trig identities;

$4+\tan x-4\sec ^2x=0$

Note that;

$\sec ^2x=\tan ^2x+1$

Input this into the last equation and we'll have;

$4+\tan x-4(\tan ^2x+1)=0$

Simplify the parenthesis;

$\begin{gathered} -4(\tan ^2x+1) \\ =-4\tan ^2x-4 \end{gathered}$

We now refine the last equation;

$\begin{gathered} 4+\tan x-4\tan ^2x-4 \\ =4-4+\tan x-4\tan ^2x \\ =\tan x-4\tan ^2x \end{gathered}$

The equation now becomes;

$\tan x-4\tan ^2x=0$

We now represent tan x by letter a.

That means;

$a-4a^2=0$

We shall apply the rule;

$\begin{gathered} \text{If} \\ ab=0 \\ \text{Then} \\ a=0,b=0 \end{gathered}$

Therefore;

$\begin{gathered} a-4a^2=0 \\ \text{Factorize;} \\ a(1-4a)=0 \end{gathered}$

At this point the solutions are;

$\begin{gathered} a=0 \\ \text{Also;} \\ 1-4a=0 \\ 1=4a \\ \frac{1}{4}=a \end{gathered}$

If we now substitute a = tan x back into the equation, we would have;

$\begin{gathered} \tan x-4\tan ^2x=0 \\ \tan x=0,\tan x=\frac{1}{4} \end{gathered}$

Where tan x = 0;

$\begin{gathered} \tan x=0 \\ x=\pi \end{gathered}$

Where tan x = 1/4;

$\begin{gathered} \tan x=\frac{1}{4} \\ x=\arctan (\frac{1}{4}) \\ x=0.24497\ldots \end{gathered}$

ANSWER:

$\begin{gathered} x=\pi \\ x=0.245 \end{gathered}$