Question

Find all solutions of each equation on the interval 0 ≤ x<2π.

Answer 1

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:

$\displaystyle xe=\begin{Bmatrix}0,1.017,2.588,\pi ,4.159,5.730\end{Bmatrix}$

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

$\displaystyle tan2x\ sec2x+2\ sec2x-tan2x=2$

Rearranging

$\displaystyle tan2x\ sec2x+2\ sec2x-tan2x-2=0$

Factoring

$\displaystyle tan2x( sec2x-1)+2 (sec2x-1)=0$

$\displaystyle (tan2x+2)(sec2x-1)=0$

We come up with two different equations:

$\displaystyle \left\{\begin{matrix}tan2x+2=0....[eq1]\\sec2x-1=0....[eq2]\end{matrix}\right.$

Let's take eq 1:

$\displaystyle tan2x=-2$

Solving for 2x

$\displaystyle 2x=arctan(-2)$

There are two sets of possible solutions:

$\left\{\begin{matrix} 2x=-1.107+2k\pi \\ 2x=-2.034+2k\pi \end{matrix}\right.$

$\displaystyle for\ k=0$

$\displaystyle \left\{\begin{matrix}2x=-1.107\\ 2x=2.034\end{matrix}\right$

We get two solutions

$\displaystyle \left\{\begin{matrix}x=-0.554\\ x=1.017\end{matrix}\right$.

The first solution is out of the range $0\leq x < 2\pi$, so it's discarded

$\displaystyle for\ k=1$

$\displaystyle \left\{\begin{matrix}2x=5.176\\ 2x=8.318\end{matrix}\right$.

$\displaystyle \left\{\begin{matrix}x=2.588\\x=4.159\end{matrix}\right$.

Both solutions are feasible

$\displaystyle for\ k=2$

$\displaystyle \displaystyle \left\{\begin{matrix}2x=11.459\\ 2x=14.601\end{matrix}\right$.

$\displaystyle \displaystyle \left\{\begin{matrix}x=5.730\\ x=7.300\end{matrix}\right$.

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:

$\displaystyle sec\ 2x-1=0$

$\displaystyle sec\ 2x=1$

This leads to the solution

$\displaystyle 2x=2k\pi$  

Or equivalently

$x=k\pi$

For k=0, x=0. This solution is valid

For k=1, $x=\pi$ . This is also valid

For k=2, $x=2\pi$ . This solution is out of range

The whole set of solutions is

$\displaystyle xe=\begin{Bmatrix}0,1.017,2.588,\pi ,4.159,5.730\end{Bmatrix}$

Answer 2

Answer:

$x = 0 \: or \:x = \pi$

Step-by-step explanation:

The given equation is

$\tan^{2} (x) \sec^{2} (x) + \sec^{2} (x) - \tan^{2} (x) = 2$

Subtract 2 from both sides and factor by grouping to get:

$\sec^{2} (x)(\tan^{2} (x) + 2) -1( \tan^{2} (x) + 2)$

$(\tan^{2} (x) + 2) (\sec^{2} (x) - 1) = 0$

By the zero product principle:

$(\tan^{2} (x) + 2) = 0 \: or \: (\sec^{2} (x) - 1) = 0$

$(\tan^{2} (x) = - 2 \: or \: \sec^{2} (x) = 1$

When

$\sec^{2} (x) = 1 \implies\cos^{2} (x) = 1$

$\implies \: \cos(x) = \pm1$

This implies

$x = 0 \: or \:x = \pi$

When

${ \tan }^{2} (x) = - 2$

We have

$\tan(x) = \pm \sqrt{ - 2}$

hence x is not defined for all real numbers