Answer
The solution set is
$$\{-0.9669+\pi n, 1.2886+\pi n, n\in Z\}$$
Work Step by Step
$$\tan x(\tan x-2)=5$$
1) Solve the equation
$$\tan x(\tan x-2)=5$$
$$\tan^2 x-2\tan x-5=0$$
Consider the equation as a quadratic formula, with $a=1, b=-2, c=-5$
- Calculate $\Delta$: $\Delta=b^2-4ac=(-2)^2-4\times1\times(-5)=4+20=24$
- Find out $\tan x$: $$\tan x=\frac{-b\pm\sqrt\Delta}{2a}=\frac{2\pm\sqrt{24}}{2}=\frac{2\pm2\sqrt6}{2}=1\pm\sqrt6$$
- For $\tan x=1+\sqrt6$. $x$ would be
$$x=\tan^{-1}(1+\sqrt6)$$
$$x\approx1.2886$$ (Be careful with degrees and radians. Here we need to use radians.)
- For $\tan x=1-\sqrt6$. $x$ would be
$$x=\tan^{-1}(1-\sqrt6)$$
$$x\approx-0.9669$$
Therefore, overall, $$x=\{-0.9669,1.2886\}$$
2) Solve the equation for all solutions
- The integer multiples of the period of the tangent function is $\pi$.
- We apply it to each solution found in step 1. The results are the solution set.
So the solution set is
$$\{-0.9669+\pi n, 1.2886+\pi n, n\in Z\}$$
