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

You can rationalize with the Weierstrass substitution, which gives
$\frac{1-{t}^{2}+2t}{1+{t}^{2}}=k\frac{2t\left(1-{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{2}}$
or ${t}^{4}-2\left(k+1\right){t}^{3}+2\left(k-1\right)t-1=0$.
The resolution of this quartic is difficult.
The method works similarly for the generalized equation.

porekalahr

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
$\mathrm{cos}x\mathrm{sin}x=\frac{1±\sqrt{{k}^{2}+1}}{{k}^{2}}$
and $\mathrm{cos}x+\mathrm{sin}x=\frac{1±\sqrt{{k}^{2}+1}}{k}$
This is a classical sum/product problem, solved with

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