Question

Solve the following equations.
$\sqrt{13-18tanx}=6tanx-3.$

Answer 1

$\sqrt{13-18tanx}=6tanx-3$

squaring both sides we get

$\Rightarrow 13-18tanx=36ta{n}^{2}x+9-36tanx$

$\Rightarrow 36ta{n}^{2}x-18tanx-4=0$

$\Rightarrow 18ta{n}^2-9tanx-2=0$

$\Rightarrow 18ta{n}^{2}x-12tanx+3tanx-2=0$

$\Rightarrow 6tanx(3tanx-2)+1(3tanx-2)=0$

$\Rightarrow (6tanx+1)(3tanx-2)=0$

$\therefore tanx=-1/6$ or $tanx=2/3$

$x=x\pi +ta{n}^{-1}(-1/6)$ or $x=ta{n}^{-1}\left(2/3\right)+m\pi$

$nϵz$ $mϵz$