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

Question

As I had solved the equation when $k=1$ in Quora and MSE by two methods, I started to investigate the equation for any real constant $k$: $$ \sin x+\cos x=k \sin x \cos x, $$

I first rewrite the equation as

$$ \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=\frac{k}{2} \sin (2 x) $$ Letting $ \displaystyle y=x-\frac{\pi}{4}$ yields $$ \begin{array}{l} \sqrt{2} \cos y=\frac{k}{2}\left(2 \cos ^{2} y-1\right) \\ 2 k \cos ^{2} y-2 \sqrt{2} \cos y-k=0 \end{array} $$ When $k\neq 0$, using quadratic formula gives

$$ \cos y=\frac{1 \pm \sqrt{1+k^{2}}}{\sqrt{2} k} $$

For real $y$, we have to restrict $\displaystyle \frac{1 \pm \sqrt{1+k^{2}}}{\sqrt{2} k}$ in $[-1,1]$. Then I found that

$$ -1 \leqslant \frac{1+\sqrt{1+k^{2}}}{\sqrt{2} k} \leqslant 1 \Leftrightarrow \quad|k| \geqslant 2 \sqrt{2} $$ and $$ -1 \leqslant \frac{1-\sqrt{1+k^{2}}}{\sqrt{2} k} \leqslant 1 \Leftrightarrow \quad k<0 \text { or } k>0 $$

Now we can conclude that

A. When $k\neq0$ $$x=n \pi-\frac{\pi}{4}$$

B. When $0\neq|k| \geqslant 2 \sqrt{2}, $ $$x=\frac{(8 n+1) \pi}{4} \pm \arccos \left(\frac{1\pm \sqrt{1+k^{2}}}{\sqrt{2} k}\right)$$

C. When $ 0 \neq|k|<2 \sqrt{2},$ $$ x= \frac{(8 n+1) \pi}{4} \pm \arccos \left(\frac{1-\sqrt{1+k^{2}}}{\sqrt{2} k}\right) $$

where $n\in Z.$

I am looking forward to seeing other methods to solve the equation.

Furthermore, how about $$a\sin x+b\cos x+c\sin x\cos x=0?$$

Answer 1

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. https://www.wolframalpha.com/input/?i=t%5E4-2%28k%2B1%29t%5E3%2B2%28k-1%29t-1%3D0

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The method works similarly for the generalized equation.

Answer 2

Let $a=\cos x$ and $b=\sin x$, then $$ \left\{\begin{array}{l} a+b=k a b \\ a^{2}+b^{2}=1 \end{array}\right. $$ Denoting the sum of $a$ and $b$ by $h$ yields $$a+b=h \textrm{ and }a b=\frac{h}{k}$$

Now we can construct a quadratic equation (*) with roots $a$ and $b$

$$ k x^{2}-h k x+h=0 \tag*{(*)} $$

Since $a$ and $b$ are roots of (*), therefore $$ \left\{\begin{array}{l} k a^{2}-h k a+h=0 \quad \cdots(1) \\ k b^{2}-h k b+h=0 \quad \cdots(2) \end{array}\right. $$ (1) + (2) yields

$$\begin{array}{l} k\left(a^{2}+b^{2}\right)-h k(a+b)+2 h=0 \\ \qquad k h^{2}-2 h-k=0 \end{array} $$

When $k\neq 0$, using quadratic formula gives

$$ \begin{aligned} h &=\frac{1 \pm \sqrt{1+k^{2}}}{k} \end{aligned} $$

By $k a b=h$, $$ \begin{aligned} \frac{k \sin 2 x}{2} &=\frac{1 \pm \sqrt{1+k^{2}}}{k} \\ \sin 2 x &=\frac{2 \pm 2 \sqrt{1+k^{2}}}{k^{2}} \end{aligned} $$

Noting that $$-1 \leqslant \frac{2+2 \sqrt{1+k^{2}}}{k^{2}} \leqslant 1 \Leftrightarrow |k| \geqslant 2 \sqrt{2}$$ $$ -1 \leqslant \frac{2-2 \sqrt{1+k^{2}}}{k^{2}} \leqslant 1 \Leftrightarrow k\neq 0, $$

we can conclude that

A. If $k=0,$

$$\displaystyle x=n \pi-\frac{\pi}{4}$$

B. If $0\neq|k| \geqslant 2 \sqrt{2}$, $$ x=\frac{1}{2}\left[n \pi+(-1)^{n} \sin ^{-1}\left(\frac{2 \pm 2 \sqrt{1+k^{2}}}{k^{2}}\right)\right] $$

C. If $0\neq |k| <2 \sqrt{2}$, $$ x=\frac{1}{2}\left[n \pi+(-1)^{n} \sin ^{-1}\left(\frac{2-2 \sqrt{1+k^{2}}}{k^{2}}\right)\right], $$ where $n \in \mathbb{Z}$.

Answer 3

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

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

Answer 4

k should be necessarily $> 1$ because each part >1.

Say since there would be 4 solutions verified for a particular numerical case of given k=2

$$ 1/\sin(t) + 1/ \cos(t) = 2, $$

we get 4 approximate solutions by iteration,.. without graphing

$$(2.81901,3.45426,1.14103,-7.55065) $$

By graphing