1.
Item 1. Determine the solution for \(2\cos^2 x-5\cos x+2=0\).
Factor the quadratic expression in terms of \(\cos x\). \(2\cos^2x-5\cos x+2=0\) \((2\cos x-1)(\cos x-2)=0\) Solve each factor separately. \(2\cos x-1=0\Rightarrow 2\cos x=1\Rightarrow \cos x=\frac{1}{2}\) \(\cos x-2=0\Rightarrow \cos x=2\), which has no solution because \(-1\leq \cos x\leq1\). So we solve \(\cos x=\frac{1}{2}\). This occurs in Quadrants I and IV with reference angle \(\frac{\pi}{3}\). \(x=\frac{\pi}{3}\) or \(x=\frac{5\pi}{3}\). The general solutions are \(x=\frac{\pi}{3}\pm2n\pi\) and \(x=\frac{5\pi}{3}\pm2n\pi\). Answer: A
2.
Item 2. Solve \(2\cos\theta\sin\theta-\sin\theta=0\), where \(0^\circ\leq\theta\leq360^\circ\).
Factor the expression. \(2\cos\theta\sin\theta-\sin\theta=0\) \(\sin\theta(2\cos\theta-1)=0\) So \(\sin\theta=0\) or \(2\cos\theta-1=0\). If \(\sin\theta=0\), then \(\theta=0^\circ,180^\circ,360^\circ\). If \(2\cos\theta-1=0\), then \(\cos\theta=\frac{1}{2}\). Cosine is positive in Quadrants I and IV. Thus \(\theta=60^\circ\) and \(300^\circ\). Combining all solutions gives \(0^\circ,60^\circ,180^\circ,300^\circ,360^\circ\). Answer: D
3.
Item 3. Determine an equivalent expression: \(\frac{\cos x}{1-\sin x}\).
This is a conjugate question. Multiply by the conjugate of the denominator. \(\frac{\cos x}{1-\sin x}\cdot\frac{1+\sin x}{1+\sin x}\) \(=\frac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)}\) \(=\frac{\cos x(1+\sin x)}{1-\sin^2x}\) Since \(1-\sin^2x=\cos^2x\), \(=\frac{\cos x(1+\sin x)}{\cos^2x}\) \(=\frac{1+\sin x}{\cos x}\) \(=\sec x(1+\sin x)\). Answer: D
4.
Item 4. Determine all restrictions for the expression: \(\frac{\sec x}{4\sin^2x-1}\).
First convert \(\sec x\) to sine and cosine. \(\frac{\sec x}{4\sin^2x-1}=\frac{\frac{1}{\cos x}}{4\sin^2x-1}\) Each denominator must not equal zero. Since \(\sec x=\frac{1}{\cos x}\), we require \(\cos x\neq0\). Also require \(4\sin^2x-1\neq0\). \(4\sin^2x\neq1\) \(\sin^2x\neq\frac{1}{4}\) \(\sin x\neq\pm\frac{1}{2}\). Therefore the restrictions are \(\cos x\neq0\) and \(\sin x\neq\pm\frac{1}{2}\). Answer: A
5.
Item 5. Determine an equivalent expression: \(\csc x-\sin x\).
Rewrite \(\csc x\) as \(\frac{1}{\sin x}\). \(\csc x-\sin x=\frac{1}{\sin x}-\sin x\) Use a common denominator. \(=\frac{1}{\sin x}-\frac{\sin^2x}{\sin x}\) \(=\frac{1-\sin^2x}{\sin x}\) Since \(1-\sin^2x=\cos^2x\), \(=\frac{\cos^2x}{\sin x}\) \(=\frac{\cos x}{\sin x}\cdot\cos x\) \(=\cot x\cos x\). Answer: A
6.
Item 6. Determine an equivalent expression: \(\sec x-\cos x\).
Rewrite \(\sec x\) as \(\frac{1}{\cos x}\). \(\sec x-\cos x=\frac{1}{\cos x}-\cos x\) Use a common denominator. \(=\frac{1}{\cos x}-\frac{\cos^2x}{\cos x}\) \(=\frac{1-\cos^2x}{\cos x}\) Since \(1-\cos^2x=\sin^2x\), \(=\frac{\sin^2x}{\cos x}\) \(=\frac{\sin x}{\cos x}\cdot\sin x\) \(=\tan x\sin x\). Answer: C
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