1.
Item 1. Solve \(\sqrt{2}\sin x+1=0\) for \(0\le x\le 2\pi\).
Solution: \(\sqrt{2}\sin x+1=0\Rightarrow \sqrt{2}\sin x=-1\Rightarrow \sin x=-\frac{1}{\sqrt2}\). The reference angle is \(\frac{\pi}{4}\). Since sine is negative in Quadrants III and IV, \(x=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\) and \(x=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\). Answer: A
2.
Item 2. Solve \(2\cos x-\sqrt3=0\) for \(0\le x\le 2\pi\).
Solution: \(2\cos x-\sqrt3=0\Rightarrow 2\cos x=\sqrt3\Rightarrow \cos x=\frac{\sqrt3}{2}\). The reference angle is \(\frac{\pi}{6}\). Since cosine is positive in Quadrants I and IV, \(x=\frac{\pi}{6}\) and \(x=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}\). Answer: C
3.
Item 3. Solve \(3\sin\theta+\sqrt3=\sin\theta\), where \(0^\circ\le \theta\le 360^\circ\).
Solution: \(3\sin\theta+\sqrt3=\sin\theta\Rightarrow 2\sin\theta+\sqrt3=0\Rightarrow \sin\theta=-\frac{\sqrt3}{2}\). The reference angle is \(60^\circ\). Since sine is negative in Quadrants III and IV, \(\theta=180^\circ+60^\circ=240^\circ\) and \(\theta=360^\circ-60^\circ=300^\circ\). Answer: B
4.
Item 4. Solve for \(x\): \(2\sin^2x-3\sin x+1=0\), \(0\le x\le 2\pi\).
Solution: Factor \(2\sin^2x-3\sin x+1=0\) as \((2\sin x-1)(\sin x-1)=0\). Thus \(\sin x=\frac12\) or \(\sin x=1\). For \(\sin x=\frac12\), \(x=\frac{\pi}{6},\frac{5\pi}{6}\). For \(\sin x=1\), \(x=\frac{\pi}{2}\). Answer: A
5.
Item 5. Solve \(4\sin\theta\cos\theta+2\sqrt2\sin\theta=2\sqrt2\cos\theta+2\), where \(0^\circ\le\theta\le360^\circ\).
Solution: Move all terms to one side and factor: \(4\sin\theta\cos\theta+2\sqrt2\sin\theta-2\sqrt2\cos\theta-2=0\). This factors as \((2\sin\theta-\sqrt2)(2\cos\theta+\sqrt2)=0\). So \(\sin\theta=\frac{\sqrt2}{2}\) or \(\cos\theta=-\frac{\sqrt2}{2}\). These give \(\theta=45^\circ,135^\circ\) from sine and \(\theta=135^\circ,225^\circ\) from cosine. The unique solutions are \(45^\circ,135^\circ,225^\circ\). Answer: A
6.
Item 6. Which expression below is equivalent to \(3\sec\frac{\pi}{7}\)?
Solution: Use the reciprocal identity \(\sec x=\frac{1}{\cos x}\). Therefore \(3\sec\frac{\pi}{7}=3\left(\frac{1}{\cos\frac{\pi}{7}}\right)=\frac{3}{\cos\frac{\pi}{7}}\). Answer: C
7.
Item 7. Simplify \(\csc\theta\tan\theta\).
Solution: \(\csc\theta=\frac{1}{\sin\theta}\) and \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Thus \(\csc\theta\tan\theta=\frac{1}{\sin\theta}\cdot\frac{\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}=\sec\theta\). Answer: D
8.
Item 8. Simplify \(\frac{\cot x}{\sec x}+\sin x\).
Solution: Convert to sine and cosine: \(\frac{\cot x}{\sec x}+\sin x=\frac{\frac{\cos x}{\sin x}}{\frac{1}{\cos x}}+\sin x=\frac{\cos^2x}{\sin x}+\sin x\). Write \(\sin x=\frac{\sin^2x}{\sin x}\), so the expression becomes \(\frac{\cos^2x+\sin^2x}{\sin x}=\frac{1}{\sin x}=\csc x\). Answer: B
9.
Item 9. Which of the following is equivalent to \(\frac{\sqrt{\csc^2x-1}}{\sqrt{\sec^2x-1}}\)?
Solution: Use \(\csc^2x-1=\cot^2x\) and \(\sec^2x-1=\tan^2x\). Then \(\frac{\sqrt{\csc^2x-1}}{\sqrt{\sec^2x-1}}=\frac{\sqrt{\cot^2x}}{\sqrt{\tan^2x}}=\frac{\cot x}{\tan x}\). Since \(\cot x=\frac{1}{\tan x}\), this becomes \(\frac{1}{\tan^2x}=\cot^2x\). Answer: B
10.
Item 10. Which expression is equivalent to \(\sin(5m)\cos(m)-\cos(5m)\sin(m)\)?
Solution: Use the sine difference identity \(\sin x\cos y-\cos x\sin y=\sin(x-y)\). Let \(x=5m\) and \(y=m\). Then \(\sin(5m)\cos(m)-\cos(5m)\sin(m)=\sin(5m-m)=\sin(4m)\). Answer: A
11.
Item 11. Simplify \(\cos(\pi+\theta)\).
Solution: Use \(\cos(x+y)=\cos x\cos y-\sin x\sin y\). Then \(\cos(\pi+\theta)=\cos\pi\cos\theta-\sin\pi\sin\theta=(-1)\cos\theta-(0)\sin\theta=-\cos\theta\). Answer: B
12.
Item 12. Identify which step shows the first mistake within this proof. Prove: \(\cos(\alpha+\beta)\times\cos(\alpha-\beta)=\cos^2\alpha-\sin^2\beta\).
Solution: The first error is Step 7. From Step 6, the expression should simplify to \(-1+\cos^2\beta+\cos^2\alpha\), not \(-1+\cos^2\beta-\cos^2\alpha\). Answer: G
13.
Item 13. Which of the following is equivalent to \(\cos^2(4x)-\sin^2(4x)\)?
Solution: Use the double-angle identity \(\cos^2\theta-\sin^2\theta=\cos(2\theta)\). Let \(\theta=4x\). Then \(\cos^2(4x)-\sin^2(4x)=\cos(8x)\). Answer: A
14.
Item 14. Write the following as a single trigonometric ratio: \(-6\sin(10x)\cos(10x)\).
Solution: Use the double-angle identity \(2\sin\theta\cos\theta=\sin(2\theta)\). Let \(\theta=10x\). Then \(2\sin(10x)\cos(10x)=\sin(20x)\). Multiply both sides by \(-3\): \(-6\sin(10x)\cos(10x)=-3\sin(20x)\). Answer: B
15.
Item 15. The following proves that \(\cos^2(2x)=1-\sin^2(2x)\). Identify the first step in which an error is made.
Solution: The first error is Step 6. Step 5 is \(2(\cos^2x+\sin^2x)-1-4\sin^2x\cos^2x\). Since \(\cos^2x+\sin^2x=1\), Step 6 should be \(2(1)-1-4\sin^2x\cos^2x\), not \(2(1)+1-4\sin^2x\cos^2x\). Answer: F
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