How to find remain factor of this trigonometic equation?

Question

The equation $$3\sin^2 x - 3\cos x -6\sin x + 2\sin 2x + 3=0$$ has a solution $x = 0$. That is mean it has a factor $\cos x - 1$. I tried write the given equation has the form $$(\cos x - 1)P(x)=0.$$ I am looking for the factor $P(x)$. How to do that?

Answer 1

$$\implies 3\left(\sin^2x-2\sin x+1\right)+2\sin2x-3\cos x=0$$

$$\implies 3\left(1-\sin x\right)^2+\cos x(4\sin x-3)=0$$

$$\text{We know,}\sin x=\frac{2t}{1+t^2},\cos2x=\frac{1-t^2}{1+t^2}\text{ where }t=\tan\frac x2$$

So, $1-\sin x=\frac{(1-t)^2}{1+t^2}, 4\sin x-3=4\frac{2t}{1+t^2}-3=-\frac{3t^2-8t+3}{1+t^2}$

$$\implies 3\left(\frac{(1-t)^2}{1+t^2}\right)^2-\frac{(1-t^2)}{1+t^2}\frac{(3t^2-8t+3)}{1+t^2}=0$$

If $t$ is finite, $$3(1-t)^4-(1-t)(1+t)(3t^2-8t+3)=0$$

$$\text{Or, }(1-t)\{3(1-t)^3-(1+t)(3t^2-8t+3)\}=0$$

If $t=1, \tan\frac x2=1\implies \frac x2=n\pi+\frac\pi4\implies x=2n\pi+\frac\pi2$ where $n$ is any integer .

Else $$2t(3t^2-7t+2)=0$$

If $t=0,t=2m\pi+0$ where $m$ is any integer (like the previous case)

Else $3t^2-7t+2=0$ (Solve the Quadratic Equation and find $x$ like the previous cases )

Finally, if $t$ is $+\infty,\frac x2=r\pi+\frac\pi2, x=2r\pi+\pi=(2r+1)\pi$ where $r$ is any integer

if $t$ is $-\infty,\frac x2=s\pi-\frac\pi2, x=2s\pi-\pi=(2s-1)\pi$ where $s$ is any integer

In the last two case $x$ is an odd multiple of $\pi$ which does not satisfy the given equation

Answer 2

It looks like the simplest way to factor out the zero-ing function is to use a half angle. For example, I rewrite your equation as

$$\sin{\frac{x}{2}} \left [ 2 \cos{\frac{x}{2}} ( 3 \sin{x} + 4 \cos{x} - 6) + 6 \sin{\frac{x}{2}} \right ] = 0$$

Note that

$$\sin{\frac{x}{2}} = \sqrt{\frac{1-\cos{x}}{2}}$$

so the $1-\cos{x}$ factor you wanted is there, in a way.

Answer 3

LHS$=sinx(3sinx-2)+4sinx(cox-1)-3(cosx-1)=\sqrt{1-cos^2 x}(3sinx-2)+4sinx(cox-1)-3(cosx-1)=\sqrt{|cosx-1||cosx+1|}(3sinx-2)+4sinx(cox-1)-3(cosx-1)$

if you insist $cosx-1$ as a factor, it has $\sqrt{|cosx-1|}$,but not $cosx-1$.

if you simply want to solve it, $1-cosx=2sin^2 \dfrac{x}{2}$,$sinx=2sin\dfrac{2}{2}cos\dfrac{x}{2}$, so more straight forward. $sin\dfrac{x}{2}$ may be a better choice.