Question

solve the trigonometric equation
$\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$

Answer 1

$\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$ is defined if $\frac{(x-3)(x^2+4)}{3x^2+4x+5}>0$

Consider $x^2+4$ has $\Delta =0-4\times 1\times 4<0$

$\therefore x^2+4>0$

Consider $3x^2+4x+5$ has $\Delta =4^2-4\times 3\times 5=16-60=-44<0$

Hence $3x^2+4x+5>0$

Consider $x-3>0\Rightarrow x\in (3,\infty )$

Hence $(3,\infty )$ is the solution.