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 \\ (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=(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=(1)/(4) \\ x=\arctan ((1)/(4)) \\ x=0.24497\ldots \end{gathered}`

ANSWER:

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