If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? 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.
 A: $(2\sin\theta+\cos\theta)^2 = 3$
$4\sin^2\theta + \cos^2\theta + 4 \sin\theta \cos\theta = 3 \sin^2\theta + 3\cos^2\theta$
$\sin^2\theta + 4\sin\theta \cos\theta = 2\cos^2\theta$
Now dividing both sides by $\cos^2\theta$,
$\tan^2\theta + 4\tan\theta = 2$
A: 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.
A: 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
;)
