Blog: Solve the equation `sin(x/2) + cos(x)

Hello!

I suppose that "sinx/2+cosx-1=0" means "sin(x/2)+cos(x)-1=0" (it can also mean "(sinx)/2+cos(x)-1=0" and "sin(x/2)+cos(x-1)=0").

To solve this equation, recall the double angle formula $\displaystyle{\cos{{\left({2}{a}\right)}}}={1}-{2}{{\sin}}^{{2}}{\left({a}\right)},$ and apply it to $\displaystyle{\cos{{\left({x}\right)}}}:$ $\displaystyle{\cos{{\left({x}\right)}}}={1}-{2}{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}.$ This way our equation becomes

$\displaystyle{\sin{{\left(\frac{{x}}{{2}}\right)}}}+{\left({1}-{2}{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}\right)}-{1}={0},$ or $\displaystyle{\sin{{\left(\frac{{x}}{{2}}\right)}}}-{2}{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}={0},$ or $\displaystyle{\sin{{\left(\frac{{x}}{{2}}\right)}}}{\left({1}-{2}{\sin{{\left(\frac{{x}}{{2}}\right)}}}\right)}={0}.$

The product is zero means at least one of factors is zero,...

Hello!

I suppose that "sinx/2+cosx-1=0" means "sin(x/2)+cos(x)-1=0" (it can also mean "(sinx)/2+cos(x)-1=0" and "sin(x/2)+cos(x-1)=0").

The product is zero means at least one of factors is zero, i.e. $\displaystyle{\sin{{\left(\frac{{x}}{{2}}\right)}}}={0}$ or $\displaystyle{\sin{{\left(\frac{{x}}{{2}}\right)}}}=\frac{{1}}{{2}}.$ These equations are well-known and their solutions are

$\displaystyle\frac{{x}}{{2}}={k}\pi,{\quad\text{or}\quad}{x}={2}{k}\pi,$
$\displaystyle\frac{{x}}{{2}}=\frac{\pi}{{6}}+{2}{k}\pi,{\quad\text{or}\quad}{x}=\frac{\pi}{{3}}+{4}{k}\pi,$
$\displaystyle\frac{{x}}{{2}}=\frac{{{5}\pi}}{{6}}+{2}{k}\pi,{\quad\text{or}\quad}{x}=\frac{{{5}\pi}}{{3}}+{4}{k}\pi,$

where $\displaystyle{k}$ is any integer.

At $\displaystyle{\left[{0},{4}\pi\right]},$ which is a period of $\displaystyle{\sin{{\left(\frac{{x}}{{2}}\right)}}}+{\cos{{\left({x}\right)}}}-{1},$ the solutions are $\displaystyle{0},$ $\displaystyle\frac{\pi}{{3}},$ $\displaystyle\frac{{{5}\pi}}{{3}},$ $\displaystyle{2}\pi$ and $\displaystyle{4}\pi.$