The Trigonometric Identity Trap: Solving an Equation with Extraneous Solutions
A student is solving the equation 2 sin^2(x) - 3 sin(x) + 1 = 0 for x in [0, 2π). They factor it as (2 sin(x) - 1)(sin(x) - 1) = 0, giving sin(x) = 1/2 or sin(x) = 1. They then write the solutions as x = π/6, 5π/6, and π/2. A classmate checks by substituting x = π/6 into the original equation and gets 2(1/4) - 3(1/2) + 1 = 0.5 - 1.5 + 1 = 0, which works. They also check x = π/2: 2(1) - 3(1) + 1 = 0, which works. However, the teacher points out that the student missed a solution. The student is confused because they thought they found all solutions in the interval. The teacher suggests graphing or using a different method to verify.
1. Identify the missing solution(s) in the interval [0, 2π) and explain why the student's method missed it.
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The missing solution is x = 5π/6. The student correctly solved sin(x)=1/2 and sin(x)=1, but for sin(x)=1/2 in the interval [0,2π), there are two solutions: x=π/6 and x=5π/6. The student only listed π/6 and π/2, forgetting the second quadrant solution. This is a common error when solving trigonometric equations: one must consider all angles in the given interval that satisfy the equation, not just the principal value. The factorization was correct, but the solution set was incomplete.
2. Solve the equation 2 sin^2(x) - 3 sin(x) + 1 = 0 using a substitution method (let u = sin(x)) and clearly state all solutions in [0, 2π). Compare your solution set with the student's and explain the discrepancy.
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Let u = sin(x). Then 2u^2 - 3u + 1 = 0. Factor: (2u-1)(u-1)=0, so u=1/2 or u=1. For u=1/2: sin(x)=1/2 ⇒ x=π/6 or x=5π/6 in [0,2π). For u=1: sin(x)=1 ⇒ x=π/2. So the complete solution set is {π/6, 5π/6, π/2}. The student's set {π/6, π/2} missed 5π/6. The discrepancy arises because the student did not find all angles in the interval for sin(x)=1/2; they only took the principal value π/6 and forgot the symmetric solution in the second quadrant.
3. Graph y = 2 sin^2(x) - 3 sin(x) + 1 on [0, 2π) and identify the x-intercepts. How does the graph confirm the number of solutions? Discuss how graphing can help avoid missing solutions in trigonometric equations.
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The graph of y = 2 sin^2(x) - 3 sin(x) + 1 on [0,2π) will cross the x-axis at x=π/6, x=5π/6, and x=π/2. Since the function is continuous and periodic, we can see three distinct intercepts in one period. Graphing helps because it provides a visual check: you can count the number of intersections with the x-axis and see if your algebraic solution set matches. It also helps identify solutions that might be overlooked, such as those in different quadrants. In this case, the graph clearly shows two intercepts for the sin(x)=1/2 case, reminding the solver to include both.