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Question 1. Solve the trigonometric equation \(\sin x-0.67=0\), for \(0\le x\le2\pi\), to two decimal places. The two solutions are:
Step 1. Rewrite the equation as \(\sin x=0.67\). Step 2. Find the reference angle: \(x=\sin^{-1}(0.67)\approx0.73\). Step 3. Since sine is positive in Quadrants I and II, the second solution is \(\pi-0.73\approx2.41\). Step 4. Therefore, the two solutions are \(0.73\) and \(2.41\). Answer: C. \(0.73,2.41\)
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Question 2. Solve \(10-3\sin x=13-7\sin x\) to two decimal places, where \(-\frac{\pi}{2}\le x\le\frac{\pi}{2}\).
Step 1. Move sine terms to one side: \(10+4\sin x=13\). Step 2. Subtract \(10\): \(4\sin x=3\). Step 3. Divide by \(4\): \(\sin x=\frac{3}{4}=0.75\). Step 4. Solve: \(x=\sin^{-1}(0.75)\approx0.85\). Step 5. This is in the given interval. Answer: A. \(0.85\)
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Question 3. Solve \(\cos\left(\frac{1}{2}x\right)=0.5\), for \(0\le x\le360^\circ\), graphically. Convert your solution to exact \(\pi\) radians. The solution is:
Step 1. Let \(u=\frac{1}{2}x\). Step 2. Solve \(\cos u=0.5\). One exact solution is \(u=\frac{\pi}{3}\). Step 3. Since \(u=\frac{x}{2}\), solve \(\frac{x}{2}=\frac{\pi}{3}\). Step 4. Multiply by \(2\): \(x=\frac{2\pi}{3}\). Answer: C. \(\frac{2\pi}{3}\)
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Question 4. Write the general solution for \(\sin\left(x-\frac{\pi}{2}\right)=0.6\). Round to four decimal places. In the answers, \(n\) is an integer.
Step 1. Let \(u=x-\frac{\pi}{2}\). Then \(\sin u=0.6\). Step 2. The reference angle is \(\sin^{-1}(0.6)\approx0.6435\). Step 3. Since sine is positive in Quadrants I and II, \(u=0.6435+2n\pi\) or \(u=\pi-0.6435+2n\pi=2.4981+2n\pi\). Step 4. Add \(\frac{\pi}{2}\) to each solution. Step 5. \(x=2.2143+2n\pi\) or \(x=4.0689+2n\pi\). Answer: D. \(2.2143+2n\pi, 4.0689+2n\pi\)
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Question 5. Solve \(\sin(x+3)=\sin x\) to four decimal places, where \(0\le x\le2\pi\). The product of the two solutions is:
Step 1. Use the identity \(\sin A=\sin B\), so \(A=B+2n\pi\) or \(A=\pi-B+2n\pi\). Step 2. The first case gives \(x+3=x+2n\pi\), which has no integer solution for \(n\). Step 3. The second case gives \(x+3=\pi-x+2n\pi\). Step 4. Solve: \(2x=\pi-3+2n\pi\), so \(x=\frac{\pi-3}{2}+n\pi\). Step 5. In \([0,2\pi]\), the two solutions are approximately \(0.0708\) and \(3.2124\). Step 6. Product: \(0.0708(3.2124)\approx0.2274\). Answer: D. \(0.2274\)
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Question 6. Solve the second-degree trigonometric equation \(\sin^2x-\frac{\sin x}{2}=0\), where \(0\le x\le2\pi\).
Step 1. Factor: \(\sin^2x-\frac{1}{2}\sin x=\sin x\left(\sin x-\frac{1}{2}\right)=0\). Step 2. Set \(\sin x=0\). On \([0,2\pi]\), this gives \(x=0,\pi,2\pi\). Step 3. Set \(\sin x=\frac{1}{2}\). On \([0,2\pi]\), this gives \(x=\frac{\pi}{6},\frac{5\pi}{6}\). Step 4. Combine all solutions. Answer: C. \(0,\frac{\pi}{6},\frac{5\pi}{6},\pi,2\pi\)
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Question 7. Solve \(\cos^2x-7\cos x+12=0\), where \(0\le x\le2\pi\).
Step 1. Let \(u=\cos x\). Then the equation becomes \(u^2-7u+12=0\). Step 2. Factor: \((u-3)(u-4)=0\). Step 3. Thus \(u=3\) or \(u=4\). Step 4. Since \(\cos x\) must be between \(-1\) and \(1\), neither value is possible. Answer: B. No solution
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Question 8. How many solutions are there for \(\sin 5x=\frac{\sqrt{2}}{2}\), where \(0\le x\le2\pi\)?
Step 1. The equation \(\sin u=\frac{\sqrt{2}}{2}\) has two solutions in every full cycle of \(u\). Step 2. Here \(u=5x\). Step 3. As \(x\) goes from \(0\) to \(2\pi\), \(5x\) goes from \(0\) to \(10\pi\), which is \(5\) full cycles. Step 4. Each cycle gives \(2\) solutions, so the total is \(5\times2=10\). Answer: A. \(10\)
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Question 9. Solve \(\cos\left(\frac{1}{2}x\right)=0\) exactly, where \(0\le x\le2\pi\).
Step 1. Let \(u=\frac{x}{2}\). Step 2. Since \(0\le x\le2\pi\), then \(0\le u\le\pi\). Step 3. Solve \(\cos u=0\) on \([0,\pi]\). This gives \(u=\frac{\pi}{2}\). Step 4. Since \(u=\frac{x}{2}\), \(\frac{x}{2}=\frac{\pi}{2}\), so \(x=\pi\). Answer: D. \(\pi\)
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Question 10. Solve \(\cos x=1-\cos x\), where \(0\le x\le2\pi\). The sum of the solutions on the domain is:
Step 1. Combine like terms: \(2\cos x=1\). Step 2. Divide by \(2\): \(\cos x=\frac{1}{2}\). Step 3. On \([0,2\pi]\), the solutions are \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\). Step 4. Sum: \(\frac{\pi}{3}+\frac{5\pi}{3}=2\pi\). Answer: C. \(2\pi\)
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Question 11. Solve \(\tan 2x=\frac{\sqrt{3}}{3}\), where \(0\le x\le2\pi\).
Step 1. \(\tan\theta=\frac{\sqrt{3}}{3}\) when \(\theta=\frac{\pi}{6}+n\pi\). Step 2. Let \(\theta=2x\), so \(2x=\frac{\pi}{6}+n\pi\). Step 3. Divide by \(2\): \(x=\frac{\pi}{12}+\frac{n\pi}{2}\). Step 4. On \([0,2\pi]\), this gives \(\frac{\pi}{12},\frac{7\pi}{12},\frac{13\pi}{12},\frac{19\pi}{12}\). Answer: D. \(\frac{\pi}{12},\frac{7\pi}{12},\frac{13\pi}{12},\frac{19\pi}{12}\)
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Question 12. Solve \(\tan^2x+1=0\), where \(0\le x\le2\pi\).
Step 1. Since \(\tan^2x\ge0\) for all values where tangent is defined, \(\tan^2x+1\ge1\). Step 2. Therefore, \(\tan^2x+1\) cannot equal \(0\). Step 3. There are no real solutions on the interval. Answer: B. No solution
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Question 13. Solve \(2\sin x=x+1\) graphically in radians to two decimal places. The solution is located at the point \((a,b)\). The value of \(a\) is:
Step 1. Graph \(y=2\sin x\) and \(y=x+1\). Step 2. The solution is the \(x\)-coordinate of the intersection. Step 3. From the graph or numerical solving, the intersection occurs at approximately \(x=-2.38\). Step 4. Therefore, \(a=-2.38\). Answer: B. \(-2.38\)
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Question 14. Use an algebraic approach to solve \(\sec(3x)=\sqrt{2}\). If \(n\) is an integer, the general solution is:
Step 1. Rewrite \(\sec(3x)=\sqrt{2}\) as \(\cos(3x)=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\). Step 2. The solutions for \(3x\) are \(3x=\frac{\pi}{4}+2n\pi\) or \(3x=\frac{7\pi}{4}+2n\pi\). Step 3. Divide by \(3\): \(x=\frac{\pi}{12}+\frac{2n\pi}{3}\) or \(x=\frac{7\pi}{12}+\frac{2n\pi}{3}\). Answer: C. \(\frac{\pi}{12}+\frac{2n\pi}{3};\frac{7\pi}{12}+\frac{2n\pi}{3}\)
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Question 15. What is the general solution to the equation \(x\sin 4x=0\)? In the answers, \(n\) is an integer.
Step 1. Use the zero product property: \(x=0\) or \(\sin 4x=0\). Step 2. The general solution for \(\sin 4x=0\) is \(4x=n\pi\). Step 3. Divide by \(4\): \(x=\frac{n\pi}{4}\). Step 4. This also includes \(x=0\) when \(n=0\). Answer: D. \(\frac{n\pi}{4}\)
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Question 16. The general solution for \(2\cos^2x+\cos x-1=0\), where \(n\) is an integer, includes: I. \(1.0472+2n\pi\), II. \(3.1416+2n\pi\), III. \(4.3798+2n\pi\), IV. \(5.2360+2n\pi\).
Step 1. Factor the equation: \(2\cos^2x+\cos x-1=(2\cos x-1)(\cos x+1)=0\). Step 2. Set \(2\cos x-1=0\), so \(\cos x=\frac{1}{2}\). This gives \(x=\frac{\pi}{3}\approx1.0472\) and \(x=\frac{5\pi}{3}\approx5.2360\). Step 3. Set \(\cos x+1=0\), so \(\cos x=-1\). This gives \(x=\pi\approx3.1416\). Step 4. The included solutions are I, II, and IV. Answer: A. I, II and IV only
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Question 17. If the identity \(\cos^2x=1-\sin^2x\) was verified graphically, then the graphs of \(y=\cos^2x\) and \(y=\sin^2x\) would:
Step 1. The functions \(\sin x\) and \(\cos x\) differ by a horizontal shift of \(\frac{\pi}{2}\). Step 2. Squaring preserves the phase relationship between \(\sin^2x\) and \(\cos^2x\). Step 3. Thus the graphs are horizontally shifted by \(\frac{\pi}{2}\) relative to each other. Answer: B. Be horizontally shifted by \(\frac{\pi}{2}\) units relative to each other
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Question 18. Which of the following are identities? I. \(\sin\left(\frac{\pi}{2}+x\right)=\sin\frac{\pi}{2}-\sin x\), II. \(\sin\left(x-\frac{\pi}{2}\right)=\sin x-\sin\frac{\pi}{2}\), III. \(\sin\left(\frac{\pi}{2}+x\right)=\cos x\), IV. \(\sin\left(\frac{\pi}{2}-x\right)=\cos x\).
Step 1. Use cofunction identities. Step 2. \(\sin\left(\frac{\pi}{2}+x\right)=\cos x\), so III is true. Step 3. \(\sin\left(\frac{\pi}{2}-x\right)=\cos x\), so IV is true. Step 4. Statements I and II incorrectly distribute sine over addition or subtraction, so they are not identities. Answer: A. III and IV
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Question 19. For what values of \(x\) is the identity \(\cot x=\frac{\cos x}{\sin x}\) defined?
Step 1. The expression \(\frac{\cos x}{\sin x}\) is undefined when \(\sin x=0\). Step 2. \(\sin x=0\) at \(x=n\pi\), where \(n\) is an integer. Step 3. Therefore, the identity is defined for all real numbers except \(x=n\pi\). Answer: A. All real numbers except \(n\pi\)
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Question 20. Which expression is equivalent to \(\frac{1+\sec x}{1+\cos x}\)?
Step 1. Rewrite \(\sec x\) as \(\frac{1}{\cos x}\). Step 2. \(\frac{1+\sec x}{1+\cos x}=\frac{1+\frac{1}{\cos x}}{1+\cos x}\). Step 3. Combine the numerator: \(1+\frac{1}{\cos x}=\frac{\cos x+1}{\cos x}\). Step 4. Divide by \(1+\cos x\): \(\frac{\frac{1+\cos x}{\cos x}}{1+\cos x}=\frac{1}{\cos x}\). Step 5. Therefore, the expression is \(\sec x\). Answer: C. \(\sec x\)
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Question 21. Which of the following is not an identity? I. \(\sec^2x-\csc^2x=\sec^2x\csc^2x\), II. \(\sec^2x+\csc^2x=\sec^2x\csc^2x\), III. \(\sec^2x+\csc^2x=(\tan x+\cot x)^2\).
Step 1. Test statement II using common denominators: \(\sec^2x+\csc^2x=\frac{1}{\cos^2x}+\frac{1}{\sin^2x}=\frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\frac{1}{\sin^2x\cos^2x}=\sec^2x\csc^2x\). So II is true. Step 2. For statement III, \((\tan x+\cot x)^2=\tan^2x+2+\cot^2x=\sec^2x+\csc^2x\), so III is true. Step 3. Statement I has subtraction instead of addition and is not true in general. Answer: C. I
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Question 22. \(\sin\left(\frac{3\pi}{2}-\frac{3\pi}{4}\right)\) can be expanded to:
Step 1. Use the sine difference identity: \(\sin(A-B)=\sin A\cos B-\cos A\sin B\). Step 2. Let \(A=\frac{3\pi}{2}\) and \(B=\frac{3\pi}{4}\). Step 3. Substitute: \(\sin\left(\frac{3\pi}{2}-\frac{3\pi}{4}\right)=\sin\frac{3\pi}{2}\cos\frac{3\pi}{4}-\cos\frac{3\pi}{2}\sin\frac{3\pi}{4}\). Answer: D. \(\sin\frac{3\pi}{2}\cos\frac{3\pi}{4}-\cos\frac{3\pi}{2}\sin\frac{3\pi}{4}\)
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Question 23. \(\cos\left(\frac{\pi}{3}+x\right)-\cos\left(\frac{\pi}{3}-x\right)\) is equivalent to:
Step 1. Use identities: \(\cos(a+x)=\cos a\cos x-\sin a\sin x\) and \(\cos(a-x)=\cos a\cos x+\sin a\sin x\). Step 2. Subtract: \(\cos(a+x)-\cos(a-x)=-2\sin a\sin x\). Step 3. Let \(a=\frac{\pi}{3}\). Then \(\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\). Step 4. Substitute: \(-2\left(\frac{\sqrt{3}}{2}\right)\sin x=-\sqrt{3}\sin x\). Answer: C. \(-\sqrt{3}\sin x\)
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Question 24. Which of the following statements are true? I. \(\cos 2x\ne2\cos x\), II. \(\cos 2x=\cos(x+x)\), III. \(\cos 2x=\cos 2+\cos x\), IV. \(2\cos x=\cos x+\cos x\).
Step 1. Statement I is true because \(\cos 2x\) is not generally equal to \(2\cos x\). Step 2. Statement II is true because \(2x=x+x\), so \(\cos 2x=\cos(x+x)\). Step 3. Statement III is false because \(\cos 2x\) does not equal \(\cos 2+\cos x\). Step 4. Statement IV is true because \(\cos x+\cos x=2\cos x\). Answer: B. I, II and IV only
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Question 25. \(\frac{2}{1-\cos 2x}\) can be simplified to:
Step 1. Use the double-angle identity \(\cos 2x=1-2\sin^2x\). Step 2. Substitute into the denominator: \(1-\cos 2x=1-(1-2\sin^2x)\). Step 3. Simplify: \(1-\cos 2x=2\sin^2x\). Step 4. Then \(\frac{2}{1-\cos 2x}=\frac{2}{2\sin^2x}=\frac{1}{\sin^2x}\). Step 5. Since \(\frac{1}{\sin^2x}=\csc^2x\), the simplified form is \(\csc^2x\). Answer: A. \(\csc^2x\)
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