Blog: How many distinct values of x satisfy the equation sin(x) = x/2, where x is in radians?

Hello!

There is one obvious solution, $\displaystyle{x}_{{1}}={0}.$ Also, because $\displaystyle{\left|{\sin{{\left({x}\right)}}}\right|}\leq{1},$ there are no solutions for $\displaystyle{\left|{x}\right|}\gt{2}.$ And one more consideration, because $\displaystyle{\sin{{\left({x}\right)}}}$ and $\displaystyle\frac{{x}}{{2}}$ are both odd functions, for each positive solution $\displaystyle{x}$ there is one negative solution $\displaystyle-{x}.$ So we can look only at $\displaystyle{x}\in{\left({0},{2}\right]}.$

Hello!

We know that $\displaystyle{\sin{{\left({x}\right)}}}$ is strictly concave down on $\displaystyle{\left({0},\frac{\pi}{{2}}\right)},$ and $\displaystyle{\sin{{\left({0}\right)}}}={0},$ $\displaystyle{\sin{{\left(\frac{\pi}{{2}}\right)}}}={1}.$ Therefore $\displaystyle{\sin{{\left({x}\right)}}}$ is greater than the secant line through $\displaystyle{\left({0},{0}\right)}$ and $\displaystyle{\left(\frac{\pi}{{2}},{1}\right)}.$ The equation of this line is $\displaystyle{y}={x}\cdot\frac{{2}}{\pi},$ so $\displaystyle{\sin{{\left({x}\right)}}}\gt{x}\cdot\frac{{2}}{\pi}>\frac{{x}}{{2}}$ for $\displaystyle{x}\in{\left({0},\frac{\pi}{{2}}\right)}.$

Well, there are no solutions on $\displaystyle{\left({0},\frac{\pi}{{2}}\right)}.$ At $\displaystyle{x}=\frac{\pi}{{2}}$ $\displaystyle{\sin{{\left({x}\right)}}}={1}\gt\frac{\pi}{{4}}=\frac{{x}}{{2}},$ at $\displaystyle{x}=\pi$ $\displaystyle{\sin{{\left({x}\right)}}}={0}\lt\frac{\pi}{{2}}=\frac{{x}}{{2}},$ so by the Intermediate Value Theorem there is at least one solution on $\displaystyle{\left(\frac{\pi}{{2}},\pi\right)}.$ We can find this solution $\displaystyle{x}_{{2}}$ only approximately, it is about 1.895.

But from $\displaystyle{x}=\frac{\pi}{{2}}$ to $\displaystyle{x}=\pi$ $\displaystyle{\sin{{\left({x}\right)}}}$ decreases and $\displaystyle\frac{{x}}{{2}}$ increases, so there is only one solution on $\displaystyle{\left(\frac{\pi}{{2}},\pi\right)}.$

So there are 3 solutions: $\displaystyle{x}_{{1}}={0},$ $\displaystyle{x}_{{2}}\approx{1.895}$ and $\displaystyle{x}_{{3}}=-{x}_{{2}}\approx-{1.895}.$