Answer:
See below.
Step-by-step explanation:
So we want to find the solutions to the two equations:
[tex]\tan(x)=-1 \text{ and } \tan(x)=5[/tex]
I)
[tex]\tan(x)=-1\\x=\tan^{-1}(-1)[/tex]
Recall the unit circle. First, note that the number inside tangent is negative. Because of this, we can be certain that the x (in radians) must be in Quadrant II and/or IV (This is because of All Students Take Calculus, where All is positive in QI, only Sine is positive in Q2, only Tangent is positive in Q3, and only Cosine is positive in QIV. Tangent is negative so the only possible choice are QII and QIV).
From the unit circle, we can see that x=3π/4 is a possible candidate since tan(3π/4)=-1.
Since tangent repeats every π, 7π/4 must also be an answer (because 3π/4 + π = 7π/4). And, as expected, 7π/4 is indeed in QIV.
Therefore, for the first equation, the solutions are:
[tex]x=3\pi/4 \text{ and } 7\pi/4[/tex]
II)
For the second equation, there is no exact value for which tangent of an angle would be equal to 5. Thus, we need to approximate.
So:
[tex]\tan(x)=5\\x=\tan^{-1}(5)\\x=\tan^{-1}(5) \text{ and } \tan^{-1}(5)+\pi[/tex]
We got the second answer because, like previously, tangent repeats every π, so we only need to add π to get the second answer.
In approximations, this is:
[tex]x\approx1.3734 \text{ and } x\approx4.5150[/tex]
Note: All the answers are in radians.
