2.
Item 2. Solve \( \tan\theta \sin\theta = \sin\theta \), where \(0^\circ \le \theta \le 360^\circ\).
Solution: \(\tan\theta\sin\theta=\sin\theta\). Move all terms to one side: \(\tan\theta\sin\theta-\sin\theta=0\). Factor: \(\sin\theta(\tan\theta-1)=0\). So \(\sin\theta=0\) or \(\tan\theta-1=0\), which means \(\tan\theta=1\). On \(0^\circ\le\theta\le360^\circ\), \(\sin\theta=0\) gives \(\theta=0^\circ,180^\circ,360^\circ\). Since tangent is positive in Quadrants I and III, \(\tan\theta=1\) gives \(\theta=45^\circ,225^\circ\). Therefore the solutions are \(0^\circ,45^\circ,180^\circ,225^\circ,360^\circ\). Answer: B
3.
Item 3. Solve \( \sin\theta + 3 = 4 \), where \(0 \le \theta \le 2\pi\).
Solution: \(\sin\theta+3=4\). Subtract \(3\) from both sides: \(\sin\theta=1\). On \(0\le\theta\le2\pi\), sine equals \(1\) only at the top of the unit circle. Therefore \(\theta=\frac{\pi}{2}\). Answer: C
4.
Item 4. Simplify \( \frac{\sin x}{\cot x} + \frac{1}{\sec x} \).
Solution: Use reciprocal and quotient identities: \(\cot x=\frac{\cos x}{\sin x}\) and \(\sec x=\frac{1}{\cos x}\). \(\frac{\sin x}{\cot x}+\frac{1}{\sec x}=\frac{\sin x}{\frac{\cos x}{\sin x}}+\frac{1}{\frac{1}{\cos x}}\). This becomes \(\sin x\left(\frac{\sin x}{\cos x}\right)+\cos x=\frac{\sin^2 x}{\cos x}+\cos x\). Write \(\cos x\) with denominator \(\cos x\): \(\frac{\sin^2 x}{\cos x}+\frac{\cos^2 x}{\cos x}=\frac{\sin^2 x+\cos^2 x}{\cos x}\). Since \(\sin^2 x+\cos^2 x=1\), the expression becomes \(\frac{1}{\cos x}=\sec x\). Answer: A
5.
Item 5. Which expression below is equivalent to \(3\sec\frac{\pi}{7}\)?
Solution: Use the reciprocal identity \(\sec x=\frac{1}{\cos x}\). Therefore \(3\sec\frac{\pi}{7}=3\left(\frac{1}{\cos\frac{\pi}{7}}\right)=\frac{3}{\cos\frac{\pi}{7}}\). Remember that the reciprocal is taken of the trig function, not of the angle. Answer: A
6.
Item 6. Determine an equivalent expression: \( \tan x+\cot x \).
Solution: Convert everything to sine and cosine: \(\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\). Use the common denominator \(\sin x\cos x\): \(\frac{\sin^2 x}{\sin x\cos x}+\frac{\cos^2 x}{\sin x\cos x}=\frac{\sin^2 x+\cos^2 x}{\sin x\cos x}\). Since \(\sin^2 x+\cos^2 x=1\), this becomes \(\frac{1}{\sin x\cos x}\). Because \(\csc x=\frac{1}{\sin x}\) and \(\sec x=\frac{1}{\cos x}\), \(\frac{1}{\sin x\cos x}=\csc x\sec x\). Answer: C
7.
Item 7. Determine an equivalent expression for \( \sin^2 4x - 1 \).
Solution: Use the Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\). Let \(\theta=4x\). Then \(\sin^2 4x+\cos^2 4x=1\). Rearrange to solve for \(\sin^2 4x-1\): \(\sin^2 4x=1-\cos^2 4x\). Therefore \(\sin^2 4x-1=(1-\cos^2 4x)-1=-\cos^2 4x\). Answer: D
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