"Solve the equation sin⁡x+cos⁡x=ksin⁡xcos⁡x for real x, where..." solved and explained

High School
Algebra
Algebra I
Quadratic function and equation

Gabriela Duarte

Answered question

2022-02-01

Solve the equation

\sin x + \cos x = k \sin x \cos x

for real x, where k is a real constant.

Answer & Explanation

gekraamdbk

Beginner2022-02-02Added 13 answers

You can rationalize with the Weierstrass substitution, which gives

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

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.

porekalahr

Beginner2022-02-03Added 10 answers

The case of $k = 0$ is elementary. Then squaring both members,
$2 \cos x \sin x + 1 = k^{2} {(\cos x \sin x)}^{2}$
immediately gives
$\cos x \sin x = \frac{1 \pm \sqrt{k^{2} + 1}}{k^{2}}$
and $\cos x + \sin x = \frac{1 \pm \sqrt{k^{2} + 1}}{k}$ .
This is a classical sum/product problem, solved with
$\cos x - \sin x = \pm \sqrt{(\cos x + \sin x)^{2} - 4 \cos x \sin x} = \pm \frac{\sqrt{(1 \pm \sqrt{k^{2} + 1})^{2} - 4 (1 \pm \sqrt{k^{2} + 1})}}{k}$

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