Why do sometimes , the answers I derive from equations not actually provide solution in inverse trignometry?

Question

This is the question:

Solve for $x$

$$\arcsin(1-x)-2\arcsin(x) = \frac{\pi}{2}$$

I solved this by these steps:

$\arcsin(1-x) = \frac{\pi}{2} + 2\arcsin(x)$

$\sin$ function on both sides

$1-x=\sin(\frac{\pi}{2} + 2\arcsin(x))$

$\implies 1-x = \cos(2\arcsin(x))$

$1-x = 1-2\sin^{2}(\arcsin(x))$

$\implies 1-x=1-2x^{2}$

Solving this we would get either $x=0$ or $x=1/2$ but when we substitute $1/2$ in the original question, it does not give out the answer of $\frac{\pi}{2}$. Why does this happen?

Answer 1

Note that $\arcsin\left(\frac12\right) = \frac\pi6.$ When you substitute $x=\frac12$ in the equation $\arcsin(1-x) = \frac{\pi}{2} + 2\arcsin(x)$, you get

$$ \arcsin\left(1 - \frac12\right) \stackrel?= \frac\pi2 + 2\arcsin\left(\frac12\right), $$

which simplifies as follows:

\begin{align} \arcsin\left(\frac12\right) &\stackrel?= \frac\pi2 + 2\arcsin\left(\frac12\right),\\ \frac\pi6 &\stackrel?= \frac\pi2 + 2\left(\frac\pi6\right),\\ \frac\pi6 &\stackrel?= \frac56\pi. \end{align}

Clearly the two sides are not equal, but it is also true that

$$ \sin\left(\frac\pi6\right) = \sin\left(\frac56\pi\right). $$

So this spurious solution was introduced when you took the sine of both sides, due to the way the sine function can give one output value for two different input values.

Answer 2

When you apply the $\sin$ function, you can create extra solutions. For a very simple example, $x=0$ has only one solution, but $\sin(x)=\sin(0)$ has infinitely many, due to the periodicity of $\sin$.

Similarly, $\arcsin(1-x)=\frac\pi2 + 2\arcsin(x)$ has one solution, but if you apply $\sin$ to both sides, you get a second solution, $x=\frac12$.

Answer 3

A careful approach...

Be careful about the domains and images (ranges) of the functions you are working with. Assuming that $x$ is a real variable (and not a complex variable), the equation $$ \arcsin(1-x) - 2\arcsin(x) = \frac{\pi}{2}, \tag{1}\label{original}$$ only makes sense if (1) $\arcsin(1-x)$ is defined, and (2) $\arcsin(x)$ is defined. Hence

1. For $\arcsin(1-x)$ to be a real number, it must be the case that $1-x \in [-1,1]$, since that is the domain of $\arcsin$. Hence $$ -1 \le 1-x \le 1 \iff -2 \le -x \le 0 \iff 2 \ge x \ge 0. $$ Thus $x \in [0,2]$. In other words, if $x$ is a solution to the original equation, then $x$ must be between $0$ and $2$ (inclusive).
2. By similar reasoning, $x \in [-1,1]$, since $\arcsin(x)$ has to make sense.

Thus far, the analysis indicates that if (\ref{original}) has any solutions, those solutions must be in the interval $$ [0,2] \cup [-1,1] = [0,1]. $$ That is, any solution must be between $0$ and $1$.

Next, observe that both of the terms on the left-hand side of (\ref{original}) must be nonnegative (since $\arcsin(y) \ge 0$ for any $y \in [0,1]$). Therefore $$ \arcsin(1-x) - 2 \arcsin(x) = \frac{\pi}{2} > 0 \implies \arcsin(1-x) > 2 \arcsin(x) > \arcsin(x). $$ But the arcsine function is strictly increasing on the interval $[-1,1]$, hence $$ \arcsin(1-x) > \arcsin(x) \implies 1-x > x \implies 1 > 2x \implies x < \frac{1}{2}. $$ Note that this inequality must be strict, since $\arcsin$ is strictly increasing. From this observation and the previous observations, it must be the case that $x \in [0,1/2)$.

Returning to (\ref{original}), \begin{align} &\arcsin(1-x) - 2\arcsin(x) = \frac{\pi}{2} \quad\text{and}\quad x \in [0,\tfrac{1}{2}) \\ &\qquad\iff \arcsin(1-x) = \frac{\pi}{2} + 2 \arcsin(x) \quad\text{and}\quad x \in [0,\tfrac{1}{2}) \\ &\qquad\iff 1-x = \sin\!\left( \frac{\pi}{2} + 2\arcsin(x) \right) = 1 - 2\sin(\arcsin(x))^2 = 1-2x^2 \quad\text{and}\quad x \in [0,\tfrac{1}{2}) \\ &\qquad\iff 2x^2 - x = (2x + 1)x = 0 \quad\text{and}\quad x \in [0,\tfrac{1}{2}) \\ &\qquad\iff x = 0. \end{align} Note that, in the last step, there are not two solutions, since it has already been determined that $x < 1/2$ (and so the second root of the equation $2x^2 - x$ cannot be a solution to the original equation).

"Spurious" solutions

In the approach to the problem presented in the question, there is an implicit assumption that $x = y$ if and only if $\sin(x) = \sin(y)$. That is, the reasoning incorrectly asserts that $$\arcsin(1-x) = \frac{\pi}{2} + 2\arcsin(x) \iff 1 - x = \sin\!\left( \frac{\pi}{2} + 2\arcsin(x) \right). $$ The "forward" implication ($\implies$) is certainly correct: if $x=y$, then it must be the case that $\sin(x) = \sin(y)$. However, the "reverse" implication ($\Longleftarrow$) doesn't generally hold. That is, there are many examples of $x$ and $y$ such that $\sin(x) = \sin(y)$, but $x \ne y$. For example, $$ \sin\!\left( \frac{\pi}{3} \right) = \sin\!\left( \frac{2\pi}{3} \right) \qquad\text{but}\qquad \frac{\pi}{3} \ne \frac{2\pi}{3}. $$ This means that when one "takes the sine" of both sides of an equation, it is possible to pick up extraneous or spurious solutions. Indeed, this is true when any non-injective function is applied to both sides of an equation: the same problem can occur when taking $n$-th roots or when applying the absolute value, for example.

Hence whenever one is working with functions that are not one-to-one, it is important to (a) know that it is possible to obtain spurious solutions, and (b) to have a strategy for eliminating those solutions. One possibility is to do as I have done, above, and very carefully work through bounds and estimates related to the domains and images of the functions being considered. This approach is, frankly, rather tedious, but has the advantage of generating an exposition which contains only "if and only if" statements.

The alternative is to work through the problem naively, and then check each possible solution at the end, eliminating the spurious solutions (as is done in the work shown in the question, and in the answers provided by David K. and 5xum). This approach has the advantage of being relatively simple, but does add an extra step to "check" the answers at the end. This final check could be rather arduous if the previous work suggests many potential solutions.

Often, both approaches should be used, in parallel: use knowledge about the behaviour of the functions to narrow down the set of possible solutions, then work through the computations naively and quickly throw away anything which doesn't live in that narrowed down set of possible solutions, and finish by checking whatever solutions remain.