Lindsay solved the equation sin x = 1 - cos x by squaring each side to obtain
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sin2 x = 1 - 2 cos x + cos2 x.
Several steps later, using correct algebra, she concluded that the solution set for solution over the interval [0, 2pi) is {0, pi/2,3pi/2}. Explain why this is not correct
Lindsay made a mistake when she squared each side of the equation sin x = 1 - cos x. Squaring an equation can introduce extraneous solutions, which are solutions that do not satisfy the original equation. In this case, squaring both sides of the equation introduces the possibility of obtaining additional solutions that are not valid.
When we square sin x = 1 - cos x, we get sin^2 x = (1 - cos x)^2. Expanding the right side of the equation gives sin^2 x = 1 - 2cos x + cos^2 x.
From this point, Lindsay continued to manipulate the equation using correct algebra. However, the resulting equation sin^2 x = 1 - 2cos x + cos^2 x does not directly help us solve for x.
To find the correct solution set, we need to go back to the original equation sin x = 1 - cos x. We can rearrange this equation to cos x = 1 - sin x and then square both sides to obtain cos^2 x = (1 - sin x)^2.
Expanding the right side gives cos^2 x = 1 - 2sin x + sin^2 x. Now we have a quadratic equation in sin x.
Rearranging the equation gives sin^2 x + 2sin x - cos^2 x + 1 = 0. This equation can be factored to (sin x + 1)(sin x - 1) = 0.
Setting each factor equal to zero gives two possible cases: sin x + 1 = 0 or sin x - 1 = 0.
For sin x + 1 = 0, we subtract 1 from both sides to obtain sin x = -1. However, there is no solution for -1 when x is restricted to the interval [0, 2pi).
For sin x - 1 = 0, we add 1 to both sides to obtain sin x = 1. The solution to this equation is x = pi/2.
Therefore, the correct solution set for the equation sin x = 1 - cos x over the interval [0, 2pi) is {pi/2}.