## Precalculus (6th Edition)

By squaring both sides, we include the case where the LHS and RHS of the initial equation are NOT EQUAL, but after squaring BECOME equal. This is the case where LHS $=a$ and RHS=$-a$. (equal in absolute value, with different signs). For example, for $x=\displaystyle \frac{3\pi}{2}$, LHS=$-1$ $RHS=+1$ so LHS$\neq$RHS, and $\displaystyle \frac{3\pi}{2}$ is not a solution. So, the solution set is incorrect. But, after squaring, we will have LHS=1=RHS. The equation produced after squaring is not equivalent to the initial equation. The solution set of the latter is not generally equal to the solution set of the former.