1. Find all solutions of each equation where $x$ may be any real number. Give the solutions in radians.
a) $\cos x=-1$
b) $\cot x=\sqrt{3}$
c) $2 \sin x-1=0$
Answer
Final Answer: \(\boxed{\text{The solutions to the equations are:}}\) a) \(x=\pi + 2k\pi\) where \(k\) is an integer, b) \(x=\frac{\pi}{6} + k\pi\) where \(k\) is an integer, and c) \(x=\frac{\pi}{6} + 2k\pi\) and \(x=\frac{5\pi}{6} + 2k\pi\) where \(k\) is an integer.
Step 1 ：Given the equations: a) \(\cos x=-1\), b) \(\cot x=\sqrt{3}\), and c) \(2 \sin x-1=0\), we need to find all solutions for \(x\) in radians.
Step 2 ：For equation a) \(\cos x=-1\), the cosine function equals -1 at \(x=\pi\) and at every odd multiple of \(\pi\). So the solutions are \(x=\pi + 2k\pi\) where \(k\) is an integer.
Step 3 ：For equation b) \(\cot x=\sqrt{3}\), the cotangent function equals \(\sqrt{3}\) at \(x=\frac{\pi}{6}\) and at every \(\pi\) interval from this point. So the solutions are \(x=\frac{\pi}{6} + k\pi\) where \(k\) is an integer.
Step 4 ：For equation c) \(2 \sin x-1=0\), solving for \(\sin x\), we get \(\sin x = \frac{1}{2}\). The sine function equals \(\frac{1}{2}\) at \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\) and at every \(2\pi\) interval from these points. So the solutions are \(x=\frac{\pi}{6} + 2k\pi\) and \(x=\frac{5\pi}{6} + 2k\pi\) where \(k\) is an integer.
Step 5 ：Final Answer: \(\boxed{\text{The solutions to the equations are:}}\) a) \(x=\pi + 2k\pi\) where \(k\) is an integer, b) \(x=\frac{\pi}{6} + k\pi\) where \(k\) is an integer, and c) \(x=\frac{\pi}{6} + 2k\pi\) and \(x=\frac{5\pi}{6} + 2k\pi\) where \(k\) is an integer.
