Question 50.
Given:
cot x = 1
Let's find the solutions of the given equation in the interval: [-2π, 2π].
To solve for the solutions, apply the following steps:
• Step 1.
Take the inverse cotangent of both sides:
[tex]\begin{gathered} x=\cot ^{-1}(1) \\ \\ x=\frac{\pi}{4} \end{gathered}[/tex]This function is positive in the first and third quadrants.
• Step 2.
To find the next solution add π to the first solution:
[tex]\begin{gathered} x=\frac{\pi}{4}+\pi \\ \\ x=\frac{\pi+4\pi}{4} \\ \\ x=\frac{5\pi}{4} \end{gathered}[/tex]• Step 3.
Find the period of cotx:
[tex]\frac{\pi}{|b|}=\frac{\pi}{|1|}=\frac{\pi}{1}=\pi[/tex]The period of the cot function will be:
[tex]x=\frac{\pi}{4}+\pi n,\text{ for any value of n.}[/tex]Substitute -2 for n and solve:
[tex]\begin{gathered} x=\frac{\pi}{4}+(-2\pi) \\ \\ x=\frac{\pi}{4}-2\pi \\ \\ x=\frac{\pi-8\pi}{4} \\ \\ x=-\frac{7\pi}{4} \end{gathered}[/tex]Substitute -1 for n:
[tex]\begin{gathered} x=\frac{\pi}{4}+\pi n \\ \\ x=\frac{\pi}{4}+(-1\pi) \\ \\ x=\frac{\pi}{4}-\pi \\ \\ x=\frac{\pi-4\pi}{4} \\ \\ x=-\frac{3\pi}{4} \end{gathered}[/tex]Substitute 0 for n:
[tex]\begin{gathered} x=\frac{\pi}{4}+\pi n \\ \\ x=\frac{\pi}{4}+0 \\ \\ x=\frac{\pi}{4} \end{gathered}[/tex]Substitute 1 for n:
[tex]\begin{gathered} x=\frac{\pi}{4}+\pi(1) \\ \\ x=\frac{5}{4} \end{gathered}[/tex]Therefore, the solutions of the equation in the interval are:
[tex]x=-\frac{7\pi}{4},-\frac{3\pi}{4},\frac{\pi}{4},\frac{5\pi}{4}[/tex]