Answer :
Answer:
[tex]x = 0 \: or \:x = \pi [/tex]
Step-by-step explanation:
The given equation is
[tex] \tan^{2} (x) \sec^{2} (x) + \sec^{2} (x) - \tan^{2} (x) = 2[/tex]
Subtract 2 from both sides and factor by grouping to get:
[tex] \sec^{2} (x)(\tan^{2} (x) + 2) -1( \tan^{2} (x) + 2)[/tex]
[tex] (\tan^{2} (x) + 2) (\sec^{2} (x) - 1) = 0[/tex]
By the zero product principle:
[tex](\tan^{2} (x) + 2) = 0 \: or \: (\sec^{2} (x) - 1) = 0[/tex]
[tex](\tan^{2} (x) = - 2 \: or \: \sec^{2} (x) = 1[/tex]
When
[tex]\sec^{2} (x) = 1 \implies\cos^{2} (x) = 1[/tex]
[tex] \implies \: \cos(x) = \pm1[/tex]
This implies
[tex]x = 0 \: or \:x = \pi [/tex]
When
[tex] { \tan }^{2} (x) = - 2[/tex]
We have
[tex] \tan(x) = \pm \sqrt{ - 2} [/tex]
hence x is not defined for all real numbers
Note: It's not clear if the expressions involving '2x' actually mean squaring the tangent/secant or doubling the angle x. I'm posting the answer assuming the latest approach.
Answer:
[tex]\displaystyle xe=\begin{Bmatrix}0,1.017,2.588,\pi ,4.159,5.730\end{Bmatrix}[/tex]
Step-by-step explanation:
Trigonometric Equations
It's a type of equations where the variable is the argument of some of the trigonometric functions. It's generally restricted to a given domain, so the solution must be iteratively selected within all the possible answers.
The equation to solve is
[tex]\displaystyle tan2x\ sec2x+2\ sec2x-tan2x=2[/tex]
Rearranging
[tex]\displaystyle tan2x\ sec2x+2\ sec2x-tan2x-2=0[/tex]
Factoring
[tex]\displaystyle tan2x( sec2x-1)+2 (sec2x-1)=0[/tex]
[tex]\displaystyle (tan2x+2)(sec2x-1)=0[/tex]
We come up with two different equations:
[tex]\displaystyle \left\{\begin{matrix}tan2x+2=0....[eq1]\\sec2x-1=0....[eq2]\end{matrix}\right.[/tex]
Let's take eq 1:
[tex]\displaystyle tan2x=-2[/tex]
Solving for 2x
[tex]\displaystyle 2x=arctan(-2)[/tex]
There are two sets of possible solutions:
[tex]\left\{\begin{matrix} 2x=-1.107+2k\pi \\ 2x=-2.034+2k\pi \end{matrix}\right.[/tex]
[tex]\displaystyle for\ k=0[/tex]
[tex]\displaystyle \left\{\begin{matrix}2x=-1.107\\ 2x=2.034\end{matrix}\right[/tex]
We get two solutions
[tex]\displaystyle \left\{\begin{matrix}x=-0.554\\ x=1.017\end{matrix}\right[/tex].
The first solution is out of the range [tex]0\leq x < 2\pi[/tex], so it's discarded
[tex]\displaystyle for\ k=1[/tex]
[tex]\displaystyle \left\{\begin{matrix}2x=5.176\\ 2x=8.318\end{matrix}\right[/tex].
[tex]\displaystyle \left\{\begin{matrix}x=2.588\\x=4.159\end{matrix}\right[/tex].
Both solutions are feasible
[tex]\displaystyle for\ k=2[/tex]
[tex]\displaystyle \displaystyle \left\{\begin{matrix}2x=11.459\\ 2x=14.601\end{matrix}\right[/tex].
[tex]\displaystyle \displaystyle \left\{\begin{matrix}x=5.730\\ x=7.300\end{matrix}\right[/tex].
Only the first solution lies in the given domain. We won't take negative values of k since it will provide negative values of x and they are not allowed in the solution
Now we solve eq 2:
[tex]\displaystyle sec\ 2x-1=0[/tex]
[tex]\displaystyle sec\ 2x=1[/tex]
This leads to the solution
[tex]\displaystyle 2x=2k\pi[/tex]
Or equivalently
[tex]x=k\pi[/tex]
For k=0, x=0. This solution is valid
For k=1, [tex]x=\pi[/tex] . This is also valid
For k=2, [tex]x=2\pi[/tex] . This solution is out of range
The whole set of solutions is
[tex]\displaystyle xe=\begin{Bmatrix}0,1.017,2.588,\pi ,4.159,5.730\end{Bmatrix}[/tex]