Lindsay solved the equation sin x = 1 - cos x by squaring each side to obtain
sin^2 x = 1 - 2 cos x + cos^2 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's mistake lies in the process of squaring both sides of the equation sin x = 1 - cos x. When squaring each side, she introduces extraneous solutions, which are solutions that are obtained from the equation manipulation but are not true solutions to the original equation.
When squaring the equation sin x = 1 - cos x, the equation becomes sin^2 x = (1 - cos x)^2. Expanding the right side gives sin^2 x = 1 - 2 cos x + cos^2 x.
However, when dealing with trigonometric equations, it is important to consider the range of the functions involved. The sine function is always between -1 and 1, while the cosine function is always between -1 and 1 as well.
By squaring the equation and expanding it, Lindsay's equation sin^2 x = 1 - 2 cos x + cos^2 x can yield additional solutions that are not actually solutions to the original equation. These additional solutions occur when sin^2 x becomes greater than 1 or less than 0, or when cos^2 x becomes greater than 1 or less than 0.
In the original equation, sin x = 1 - cos x, the only valid solutions within the interval [0, 2pi) that satisfy the equation are x = 0, x = pi/2, and x = 3pi/2. However, by squaring both sides and introducing extraneous solutions, Lindsay mistakenly added additional solutions that do not satisfy the original equation, leading to an incorrect solution set.