2.
Item 2. Solve \(2\sin x-1=0\) for \(0\le x
Solution: \(2\sin x-1=0\Rightarrow 2\sin x=1\Rightarrow \sin x=\frac12\). The reference angle is \(\frac{\pi}{6}\). Since sine is positive in Quadrants I and II, \(x=\frac{\pi}{6}\) and \(x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\). Answer: A
3.
Item 3. Solve \(2\cos x+\sqrt3=0\) for \(0\le x
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 negative in Quadrants II and III, \(x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\) and \(x=\pi+\frac{\pi}{6}=\frac{7\pi}{6}\). Answer: A
4.
Item 4. Solve \(\tan\theta\sin\theta=\sin\theta\), where \(0^\circ\le\theta\le360^\circ\).
Solution: \(\tan\theta\sin\theta=\sin\theta\Rightarrow \tan\theta\sin\theta-\sin\theta=0\). Factor: \(\sin\theta(\tan\theta-1)=0\). Thus \(\sin\theta=0\) or \(\tan\theta=1\). On \(0^\circ\le\theta\le360^\circ\), \(\sin\theta=0\) gives \(0^\circ,180^\circ,360^\circ\), and \(\tan\theta=1\) gives \(45^\circ,225^\circ\). Answer: D
5.
Item 5. Solve \(\tan^2x=\tan x\), \(0\le x\le360^\circ\).
Solution: \(\tan^2x=\tan x\Rightarrow \tan^2x-\tan x=0\). Factor: \(\tan x(\tan x-1)=0\). Thus \(\tan x=0\) or \(\tan x=1\). \(\tan x=0\) gives \(x=0,\pi\). \(\tan x=1\) gives \(x=\frac{\pi}{4},\frac{5\pi}{4}\). Therefore \(x=0,\frac{\pi}{4},\pi,\frac{5\pi}{4}\). Answer: B
6.
Item 6. Solve \(\sin\theta\cos\theta+\cos\theta=-\sin\theta-1\), where \(0\le\theta\le2\pi\).
Solution: \(\sin\theta\cos\theta+\cos\theta=-\sin\theta-1\). Factor the left side and right side: \(\cos\theta(\sin\theta+1)=-(\sin\theta+1)\). Move all terms to one side: \(\cos\theta(\sin\theta+1)+(\sin\theta+1)=0\). Factor: \((\sin\theta+1)(\cos\theta+1)=0\). Thus \(\sin\theta=-1\) or \(\cos\theta=-1\). Therefore \(\theta=\frac{3\pi}{2}\) or \(\theta=\pi\). Answer: B
7.
Item 7. Simplify \(\csc\theta\tan\theta\).
Solution: Use \(\csc\theta=\frac{1}{\sin\theta}\) and \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Then \(\csc\theta\tan\theta=\frac{1}{\sin\theta}\cdot\frac{\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}=\sec\theta\). Answer: B
8.
Item 8. Determine all restrictions for the expression \(\frac{\sec x}{4\sin^2x-1}\).
Solution: Since \(\sec x=\frac{1}{\cos x}\), we need \(\cos x\ne0\). Also the denominator \(4\sin^2x-1\) cannot be zero. \(4\sin^2x-1\ne0\Rightarrow 4\sin^2x\ne1\Rightarrow \sin^2x\ne\frac14\Rightarrow \sin x\ne\pm\frac12\). Answer: C
9.
Item 9. Determine an equivalent expression: \(\tan x+\cot x\).
Solution: Convert to sine and cosine: \(\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\). Use the common denominator \(\sin x\cos x\): \(\frac{\sin^2x+\cos^2x}{\sin x\cos x}=\frac{1}{\sin x\cos x}\). Since \(\csc x=\frac1{\sin x}\) and \(\sec x=\frac1{\cos x}\), this equals \(\csc x\sec x\). Answer: C
10.
Item 10. Determine an equivalent expression: \(\sec x-\cos x\).
Solution: \(\sec x-\cos x=\frac{1}{\cos x}-\cos x=\frac{1-\cos^2x}{\cos x}\). Since \(1-\cos^2x=\sin^2x\), this becomes \(\frac{\sin^2x}{\cos x}=\frac{\sin x}{\cos x}\cdot\sin x=\tan x\sin x\). Answer: C
11.
Item 11. Which of the following can be written as \(\sin(9x)\)?
Solution: Use the sine sum identity \(\sin a\cos b+\cos a\sin b=\sin(a+b)\). With \(a=5x\) and \(b=4x\), \(\sin(5x)\cos(4x)+\cos(5x)\sin(4x)=\sin(9x)\). Answer: A
12.
Item 12. Which expression is equivalent to \(\cos(3m)\cos(2m)+\sin(3m)\sin(2m)\)?
Solution: Use the cosine difference identity \(\cos x\cos y+\sin x\sin y=\cos(x-y)\). Let \(x=3m\) and \(y=2m\). Then \(\cos(3m)\cos(2m)+\sin(3m)\sin(2m)=\cos(3m-2m)=\cos(m)\). Answer: B
13.
Item 13. Identify which step is incorrect. Prove: \(\sin(\alpha+\beta)\times\sin(\alpha-\beta)=\sin^2\alpha-\sin^2\beta\).
Solution: The first incorrect step is Step 4. The expression should be \(\sin^2\alpha(1-\sin^2\beta)-(1-\sin^2\alpha)\sin^2\beta\), but the proof uses a plus sign instead. Answer: D
14.
Item 14. Simplify \(\frac{2\cos\theta}{\sin 2\theta}\).
Solution: Use \(\sin2\theta=2\sin\theta\cos\theta\). Then \(\frac{2\cos\theta}{\sin2\theta}=\frac{2\cos\theta}{2\sin\theta\cos\theta}=\frac1{\sin\theta}=\csc\theta\). Answer: B
15.
Item 15. \(1-\sin^2\left(\frac{\pi}{6}\right)\) is equivalent to:
Solution: Use the cosine double-angle identity \(\cos2x=1-2\sin^2x\). With \(x=\frac{\pi}{6}\), \(1-2\sin^2\left(\frac{\pi}{6}\right)=\cos\left(\frac{\pi}{3}\right)\). Therefore the matching option is \(\cos\left(\frac{\pi}{3}\right)\). Answer: D
16.
Item 16. Identify which step is incorrect. Prove: \(3\cos^2\theta+2\cos(2\theta)=5-7\sin^2\theta\).
Solution: The proof is correct. There is no error. Answer: E
1 out of 1
