2.
Item 2. Solve \(\sqrt{2}\sin x+1=0\) for \(0\le x
Solution: Start with \(\sqrt{2}\sin x+1=0\). Subtract \(1\): \(\sqrt{2}\sin x=-1\). Divide by \(\sqrt{2}\): \(\sin x=-\frac{1}{\sqrt{2}}\). The reference angle is \(\frac{\pi}{4}\). Since sine is negative in Quadrants III and IV, \(x_1=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\) and \(x_2=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\). Answer: B
3.
Item 3. Determine the general solutions for \(2\sin x\cos x-\cos x=0\), giving exact answers where appropriate.
Solution: Factor the expression: \(2\sin x\cos x-\cos x=0\) becomes \(\cos x(2\sin x-1)=0\). Solve each factor separately. From \(\cos x=0\), \(x=\frac{\pi}{2},\frac{3\pi}{2}\). From \(2\sin x-1=0\), we get \(2\sin x=1\), so \(\sin x=\frac{1}{2}\). Sine is positive in Quadrants I and II with reference angle \(\frac{\pi}{6}\), so \(x=\frac{\pi}{6},\frac{5\pi}{6}\). Therefore the general solutions are \(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{\pi}{2},\frac{3\pi}{2}+2\pi n\). Answer: B
4.
Item 4. Which expression is equivalent to \(\frac{\cos x}{\sec x}-\cos x\sec x\)?
Solution: Use \(\sec x=\frac{1}{\cos x}\). \(\frac{\cos x}{\sec x}-\cos x\sec x=\frac{\cos x}{\frac{1}{\cos x}}-\cos x\left(\frac{1}{\cos x}\right)\). This simplifies to \(\cos^2 x-1\). Using \(\sin^2 x+\cos^2 x=1\), we have \(\cos^2 x-1=-\sin^2 x\). Answer: C
5.
Item 5. Which expression below is equivalent to \(2\cot\frac{\pi}{5}\)?
Solution: Use the reciprocal identity \(\cot x=\frac{1}{\tan x}\). Therefore \(2\cot\frac{\pi}{5}=2\left(\frac{1}{\tan\frac{\pi}{5}}\right)=\frac{2}{\tan\frac{\pi}{5}}\). Remember that the reciprocal is taken of the trig function, not the angle. Answer: D
6.
Item 6. Simplify \(\frac{\cot x}{\sec x}+\sin x\).
Solution: Rewrite the trig functions using sine and cosine: \(\cot x=\frac{\cos x}{\sin x}\) and \(\sec x=\frac{1}{\cos x}\). Then \(\frac{\cot x}{\sec x}+\sin x=\frac{\frac{\cos x}{\sin x}}{\frac{1}{\cos x}}+\sin x\). This becomes \(\frac{\cos^2 x}{\sin x}+\sin x\). Write \(\sin x\) with denominator \(\sin x\): \(\frac{\cos^2 x}{\sin x}+\frac{\sin^2 x}{\sin x}=\frac{\cos^2 x+\sin^2 x}{\sin x}\). Since \(\cos^2 x+\sin^2 x=1\), the expression becomes \(\frac{1}{\sin x}=\csc x\). Answer: B
7.
Item 7. Which of the following is equivalent to \(\frac{\sqrt{\sec^2 x-1}}{\sqrt{1-\cos^2 x}}\)?
Solution: Use the identities \(\sec^2 x-1=\tan^2 x\) and \(1-\cos^2 x=\sin^2 x\). \(\frac{\sqrt{\sec^2 x-1}}{\sqrt{1-\cos^2 x}}=\frac{\sqrt{\tan^2 x}}{\sqrt{\sin^2 x}}\). This simplifies to \(\frac{\tan x}{\sin x}\). Since \(\tan x=\frac{\sin x}{\cos x}\), \(\frac{\tan x}{\sin x}=\frac{\frac{\sin x}{\cos x}}{\sin x}=\frac{1}{\cos x}=\sec x\). Answer: A
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