Trigonometry equation $\tan2x+3 \sec x +3=0$ for $0\leq x\leq 360$ Solve the trigonometry equation for $0\leq x\leq 360$

$$\tan2x+3 \sec x +3=0$$

I've no idea how should I start. Is there any identity I've to use? Hope someone can show out the working and explain for it. Thanks in advance.
Clearly, $\cos2x\cos x\ne0$

Multiply both sides of the given equation $$\cos x\sin2x+3(1+\cos x)\cos2x=0$$ 

$$0=2\sin x\cos^2x+3(1+\cos x)\cos2x$$   

$$=4\sin\dfrac x2\cos\dfrac x2\cos^2x+3\cdot2\cos^2\dfrac x2\cos2x$$

$$=2\cos\dfrac x2\left(2\sin\dfrac x2\cos^2x+3\cos\dfrac x2\cos2x\right)$$

If $\cos\dfrac x2=0,\dfrac x2=(2n+1)90^\circ\implies x=(2n+1)180^\circ\equiv180^\circ\pmod{360^\circ}$

Else  $2\sin\dfrac x2\cos^2x+3\cos\dfrac x2\cos2x=0$

$-2\tan\dfrac x2=\dfrac{3(2\cos^2x-1)}{\cos^2x}$

Use $\cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$ which unfortunately leaves us with a bi-quadratic equation in $\tan\dfrac x2$
Hint in $[0,\pi]$.

let $f(x)=\tan(2x)+\frac{3}{\cos(x)}+3$

$$f'(x)=2(1+\tan^2(2x))+\frac{3\sin(x)}{\cos^2(x)}$$

$f$ is continuous and strictly increasing in $[0,\frac{\pi}{4})$ ,$(\frac{\pi}{4},\frac{\pi}{2})$, $(\frac{\pi}{2},\frac{3\pi}{4})$ and $(\frac{3\pi}{4},\pi]$

so in $(\frac{3\pi}{4},\pi]$, the only solution is

$x=\pi$.