If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$?

Question

If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?

$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$

First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neither of these known angles worked. But by doing it I realized that for $\theta=\frac{\pi}4+k\pi\quad$, $\tan^2\theta+4\tan\theta=5\quad$ and $2\sin\theta+\cos\theta\neq\sqrt3$ Hence the fourth choice is wrong.

Also tried to expanding, $$\tan^2\theta+4\tan\theta=\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{4\sin\theta}{\cos\theta}=\dfrac{\sin^2\theta+4\sin\theta\cos\theta}{\cos^2\theta}$$But can't continue even writing $4\sin\theta\cos\theta=2\sin2\theta$ doesn't help.

Answer 1

$(2\sin\theta+\cos\theta)^2 = 3$

$4\sin^2\theta + \cos^2\theta + 4 \sin\theta \cos\theta = 3 \sin^2\theta + 3\cos^2\theta$

$\sin^2\theta + 4\sin\theta \cos\theta = 2\cos^2\theta$

Now dividing both sides by $\cos^2\theta$,

$\tan^2\theta + 4\tan\theta = 2$

Answer 2

Alternative: We have $$2\sin\theta=\sqrt3 -\cos \theta$$ Squaring, $$4-4\cos^2 \theta=3+\cos^2\theta-2\sqrt 3 \cos\theta$$ So, $$5\cos^2 \theta-2\sqrt 3 \cos\theta-1=0$$ Or $$\cos \theta=\frac {2\sqrt 3 \pm 4\sqrt 2}{10}$$ Since further conditions haven't been given, both values would be valid, and from here $\tan \theta$ can be calculated.

Answer 3

Simply square the constraint and you get $4 \sin^2(x)+4\sin(x)\cos(x)+\cos^2(x)=3$ then double angle and trig identity $3\sin^2(x)+2\sin(2x)=2$ furthermore this is equivalent to $2\sin(2x) = 3\cos^2(x)-1$

and with your definition of the left hand side

$\tan^2(x)+4\tan(x)=\frac{\sin^2(x)+2\sin(2x)}{\cos^2(x)}$

we can insert the definition of $2\sin(2x)$ we got from the constraint and we get

$\tan^2(x)+4\tan(x)=\frac{\sin^2(x)+3\cos^2(x)-1}{\cos^2(x)}=\frac{2\cos^2(x)}{\cos^2(x)}=2$

I hope that helps and I hope I didn't make a mistake ;)