Gabriela Duarte

2022-02-01

Solve the equation $\mathrm{sin}x+\mathrm{cos}x=k\mathrm{sin}x\mathrm{cos}x$ for real x, where k is a real constant.

gekraamdbk

Beginner2022-02-02Added 13 answers

You can rationalize with the Weierstrass substitution, which gives

$\frac{1-{t}^{2}+2t}{1+{t}^{2}}=k\frac{2t(1-{t}^{2})}{{(1+{t}^{2})}^{2}}$

or${t}^{4}-2(k+1){t}^{3}+2(k-1)t-1=0$ .

The resolution of this quartic is difficult.

The method works similarly for the generalized equation.

or

The resolution of this quartic is difficult.

The method works similarly for the generalized equation.

porekalahr

Beginner2022-02-03Added 10 answers

The case of $k=0$ is elementary. Then squaring both members,

$2\mathrm{cos}x\mathrm{sin}x+1={k}^{2}{\left(\mathrm{cos}x\mathrm{sin}x\right)}^{2}$

immediately gives

$\mathrm{cos}x\mathrm{sin}x=\frac{1\pm \sqrt{{k}^{2}+1}}{{k}^{2}}$

and $\mathrm{cos}x+\mathrm{sin}x=\frac{1\pm \sqrt{{k}^{2}+1}}{k}$.

This is a classical sum/product problem, solved with

$\mathrm{cos}x-\mathrm{sin}x=\pm \sqrt{(\mathrm{cos}x+\mathrm{sin}x{)}^{2}-4\mathrm{cos}x\mathrm{sin}x}=\pm \frac{\sqrt{(1\pm \sqrt{{k}^{2}+1}{)}^{2}-4(1\pm \sqrt{{k}^{2}+1})}}{k}$