Thread: Soving equation involving trigo. func.

1. Soving equation involving trigo. func.

Solve, for $\displaystyle 0<x<2\pi$, the equation $\displaystyle 2cos^2x+cosxsinx-6sin^2x=0$

I tried to divide both sides by $\displaystyle cos^2x$ and obtain
$\displaystyle 2+tanx-6tan^2x=0$, which can be solved easily.

I can only divide both sides by $\displaystyle cos^2x$ provided that it is non-zero, right?

Do I have to check whether $\displaystyle cos^2x=0$ will give some other solution to the equation??

Thanks!!

2. If $\displaystyle \cos x=0$ then $\displaystyle \sin x=0$, but $\displaystyle \sin x$ and $\displaystyle \cos x$ can't be 0 simultaneously.
So you can divide the equation by $\displaystyle \cos^2x$ without loosing any solution.

3. Hello, acc100jt!

Another approach . . .

Solve, for $\displaystyle 0 <x<2\pi$, the equation $\displaystyle 2\cos^2\!x+\cos x\sin x-6\sin^2\!x\:=\:0$

Factor: .$\displaystyle (2\cos x - 3\sin x)(\cos x + 2\sin x) \:=\:0$

And solve the two equations . . .

$\displaystyle 2\cos x \,- \,3\sin x \;=\;0\quad\Rightarrow\quad3\sin x\:=\:2\cos x\quad\Rightarrow\quad\frac{\sin x}{\cos x} \,=\,\frac{2}{3}$
. . $\displaystyle \tan x \,=\,\frac{2}{3}\quad\Rightarrow\quad x \,=\,\tan^{-1}\left(\frac{2}{3}\right)$

$\displaystyle \cos x + 2\sin x\;=\;0\quad\Rightarrow\quad 2\sin x \,=\,\text{-}\cos x\quad\Rightarrow\quad\frac{\sin x}{\cos x} \,=\,\text{-}\frac{1}{2}$
. . $\displaystyle \tan x \,=\,\text{-}\frac{1}{2}\quad\Rightarrow\quad x \:=\:\tan^{-1}\left(\text{-}\frac{1}{2}\right)$

. . . with the same results.