1.
Item 1. Solve \(2\cos^2x-1=0\), for \(0\leq x\leq 2\pi\).
First isolate \(\cos x\). \(2\cos^2x-1=0\) \(2\cos^2x=1\) \(\cos^2x=\frac{1}{2}\) \(\cos x=\pm\frac{1}{\sqrt{2}}\) The reference angle is \(\frac{\pi}{4}\). Since cosine is both positive and negative, solutions occur in all four quadrants. Quadrant I: \(x=\frac{\pi}{4}\) Quadrant II: \(x=\pi-\frac{\pi}{4}=\frac{3\pi}{4}\) Quadrant III: \(x=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\) Quadrant IV: \(x=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\) Answer: A
2.
Item 2. Solve \(2\sin x-1=0\) for \(0\leq x
First isolate \(\sin x\). \(2\sin x-1=0\) \(2\sin x=1\) \(\sin x=\frac{1}{2}\) Sine is positive in Quadrants I and II. The reference angle for \(\sin x=\frac{1}{2}\) is \(\frac{\pi}{6}\). Quadrant I: \(x=\frac{\pi}{6}\) Quadrant II: \(x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\) Answer: C
3.
Item 3. Simplify \(\cot x\sec x\).
Rewrite each expression using sine and cosine. \(\cot x=\frac{\cos x}{\sin x}\) \(\sec x=\frac{1}{\cos x}\) \(\cot x\sec x=\left(\frac{\cos x}{\sin x}\right)\left(\frac{1}{\cos x}\right)\) The \(\cos x\) factors cancel. \(=\frac{1}{\sin x}\) \(=\csc x\) Answer: B
4.
Item 4. Simplify \(\frac{\sin x}{\cot x}+\frac{1}{\sec x}\).
Rewrite \(\cot x\) and \(\sec x\) using sine and cosine. \(\cot x=\frac{\cos x}{\sin x}\) and \(\sec x=\frac{1}{\cos x}\). \(\frac{\sin x}{\cot x}+\frac{1}{\sec x}\) \(=\frac{\sin x}{\frac{\cos x}{\sin x}}+\frac{1}{\frac{1}{\cos x}}\) \(=\sin x\left(\frac{\sin x}{\cos x}\right)+\cos x\) \(=\frac{\sin^2x}{\cos x}+\cos x\) Use a common denominator. \(=\frac{\sin^2x}{\cos x}+\frac{\cos^2x}{\cos x}\) \(=\frac{\sin^2x+\cos^2x}{\cos x}\) Since \(\sin^2x+\cos^2x=1\), \(=\frac{1}{\cos x}=\sec x\) Answer: C
5.
Item 5. Determine an equivalent expression: \(\tan x+\cot x\).
Convert everything to sine and cosine. \(\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\) Use a common denominator. \(=\frac{\sin^2x}{\sin x\cos x}+\frac{\cos^2x}{\sin x\cos x}\) \(=\frac{\sin^2x+\cos^2x}{\sin x\cos x}\) Since \(\sin^2x+\cos^2x=1\), \(=\frac{1}{\sin x\cos x}\) \(=\csc x\sec x\) Answer: B
6.
Item 6. Which of the following is equivalent to \(\frac{\sqrt{\sec^2x-1}}{\sqrt{1-\cos^2x}}\)?
Use the identity \(\sec^2x-1=\tan^2x\). Also use the identity \(1-\cos^2x=\sin^2x\). \(\frac{\sqrt{\sec^2x-1}}{\sqrt{1-\cos^2x}}=\frac{\sqrt{\tan^2x}}{\sqrt{\sin^2x}}\) \(=\frac{\tan x}{\sin x}\) Rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\). \(=\frac{\frac{\sin x}{\cos x}}{\sin x}\) \(=\frac{1}{\cos x}\) \(=\sec x\) Answer: B
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