Sorry for troubling you guys but i keep coming up with an extra 3cosX
Prove:
3sinxtanx=8 is the same as 3cos^2x+8cosx-3=0
Any help would be great.
$\displaystyle tan{x} = \frac{sin(x)}{cos(x)}$
$\displaystyle \frac{3sin^2(x)}{cos(x)}-8 = 0$
$\displaystyle \frac{3-3cos^2(x)}{cos(x)}-8=0$
$\displaystyle -8 = -8\frac{cos(x)}{cos(x)}$
$\displaystyle \frac{3-3cos^2(x)}{cos(x)}-8\frac{cos(x)}{cos(x)}=0$
$\displaystyle \frac{3-3cos^2(x)-8cos(x)}{cos(x)} = 0$
Then cos(x) will disappear and take everything to the other side to get the answer
Hi Dominic,
I'm not sure if this is the proof you are looking for, but here is something for what it's worth.
$\displaystyle 3 \cos^2x+8 \cos x -3=0$
$\displaystyle (3 \cos x-1)(\cos x +3)=0$
$\displaystyle 3\cos x-1=0$
$\displaystyle \cos x={\color{red}\frac{1}{3}}$
Now, use that in the 1st equation.
$\displaystyle 3 \sin x \tan x = 8$
$\displaystyle 3 \sin x \left(\frac{\sin x}{\cos x}\right)=8$
$\displaystyle \frac{3 \sin^2x}{\cos x}=8$
$\displaystyle \frac{3 \sin^2 x}{{\color{red}\frac{1}{3}}}=8$
$\displaystyle 9 \sin^2x=8$
$\displaystyle \sin^2x=\frac{8}{9}$
$\displaystyle \sin x=\frac{2\sqrt{2}}{3}$
Therefore, $\displaystyle \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)=\cos^{-1} \left(\frac{1}{3}\right)$
Hello, Dominic O'L
Prove: .$\displaystyle 3\sin x\tan x\:=\:8$ is the same as $\displaystyle 3\cos^2\!x+ 8\cos x -3\:=\:0$
We have: .$\displaystyle 3\sin x\cdot\frac{\sin x}{\cos x} \:=\:8 \quad\Rightarrow\quad \frac{3\sin^2\!x}{\cos x} \:=\:8 \quad\Rightarrow\quad 3\sin^2\!x \:=\:8\cos x$
. . $\displaystyle 3(1-\cos^2\!x) \:=\:8\cos x \quad\Rightarrow\quad 3-3\cos^2\!x \:=\: 8\cos x$
Therefore: .$\displaystyle \boxed{3\cos^2\!x + 8\cos x - 3 \:=\:0}$