How to solve $\cos x-x=-\frac{\pi}2$ without using the calculator

Question

I feel like this is a really simple question, but I can't seem to find the answer, and I'm just stuck on how to even start.

$$\cos x-x=-\frac{\pi}2$$

What's the first thing I should do?

Answer 1

This is not such a simple question because you are deemed to find all solutions. By inspection you might find that $x=\frac\pi2$ is one.

Now let us consider the function $\cos x-x+\frac\pi2$ and find its extrema. By differentiation,

$$-\sin x-1=0\iff x=\frac{3\pi}2+2k\pi.$$

The corresponding values of the function are

$$f_k=(2k+1)\pi$$

and there is a single change of sign, between $-\frac\pi2$ ($k=-1$) and $\frac{3\pi}2$ ($k=0$).

Answer 2

You're trying to solve for $x$? In principle such an equation involving both a trig function and a non-constant polynomial will generally have to be solved by numerical methods. As such if you see one and are asked to solve it exactly, you should try guessing some particular "nice" values to see if they work. After that you should think a little bit about whether you have found all solutions.

Answer 3

It's $\pi / 2$ as can be seen by trying. Note that it's one possible solution and as has been pointed out, generally there could be more. But without means to evaluate the involved function $cos$ at all points, you have to rely on intuition and trying.

Answer 4

In order to oslve $$ \cos x = x-\frac \pi 2$$ substitue $x\leftarrow y+\frac \pi2$. This makes $\cos x=\cos(y+\tfrac\pi 2)=-\sin y$, so we want to solve $$ -\sin y = y.$$ Bot sides are odd functions of $y$, hence $y=0$ is certainly a solution. There is no positive (and by symmetry of odd functions: also no negative) solution because $$-\sin y<0<y\qquad \text{for }0<y<\pi $$ and $$ -\sin y\le 1<y\qquad\text{for }y>1.$$ It follows that $$x=\frac\pi2$$ is the only solution of the original equation.