Answer
Linda squared each side, but when she came up with the solution set, she did not eliminate the extraneous solution, which is $\frac{3\pi}{2}$.
Therefore, the correct solution set is only $\{0,\frac{\pi}{2}\}$
Work Step by Step
1) Summarize the exercise:
- The equation is $\sin x=1-\cos x$
- The equation should be solved over the interval $[0,2\pi)$
- Linda squared each side to get $$\sin^2x=1-2\cos x+\cos^2x$$
- Linda did CORRECT ALGEBRA and ended up with the solution set $\{0,\frac{\pi}{2},\frac{3\pi}{2}\}$
2) Examine each possibility of problems:
- The exercise states that Linda did correct algebra, so the problem does not lie in calculations.
- All the solutions in the solution set lie in the interval $[0,2\pi)$, so this is not a problem of false solutions outside the interval.
- That means the only problem left should lie in the squaring step.
Normally when we solve an arbitrary equation like this: $$x=A$$
Use the squaring method: $$x^2=A^2$$
After some algebraic steps, the aim is still to find $x$, so we need to bring $x^2$ back to first degree, yet now $$x=\pm A$$
So as you see, now we have an extra solution $x=-A$, which is extraneous since we care only about original equation $x=A$.
Therefore, to eliminate the extraneous solutions, we need to try each found solution back in the original equation.
3) Back to main exercise: $$\sin x=1-\cos x$$
- For $x=0$, $1-\cos0=1-1=0=\sin0$. So $0$ is a correct solution.
- For $x=\frac{\pi}{2}$, $1-\cos\frac{\pi}{2}=1-0=1=\sin\frac{\pi}{2}$. $\frac{\pi}{2}$ is also a correct solution.
- For $x=\frac{3\pi}{2}$, $1-\cos\frac{3\pi}{2}=1-0=1\ne\sin\frac{3\pi}{2}$, since $\sin\frac{3\pi}{2}=-1$. So $\frac{3\pi}{2}$ is not a correct solution.
Therefore, the solution set is only $\{0,\frac{\pi}{2}\}$
