Solve the equation for $4\csc (x+2)-3\cot^2(x+2) = 3$ for $ 0\leq x \leq 8$ For this question, what I did was simply change all of the trigonometric functions to $\sin , \cos, \tan$ form. So the final step before I got stuck was $3(\sin^2 (x+2) + \cos^2 (x+2))=4\sin(x+2)$.
Not quite sure how to proceed from here, any hint will be appreciated.
 A: Let's put $\theta = x+2$, work from there first. Then $$0\leq x\leq 8 \iff 2\leq \theta \leq 10$$
Then we we have $$3(\underbrace{\sin^2 \theta +\cos^2}_{ \large = 1}) = 4\sin \theta$$
$$ =  3(1) = 4\sin \theta \iff \sin\theta= \frac 34, \quad \forall\, \theta \in [2, 10]$$
We can solve for \theta, by using the inverse sin function  on both sides of the equation: $$\sin^{-1}(\sin \theta)  = \sin^{-1}\left(\frac 34\right)$$ 
$$\theta =\sin^{-1}\left(\frac 34\right)$$
I'll let you determine what solutions exist, for $2\leq \theta \leq 10$

Note, we can also back substitute $x+2 = \theta$ to get $$x+2 = \sin^{-1}\left(\frac 34\right) \iff x= \sin^{-1}\left(\frac 34\right) -2$$ but by back substitution, we also need to specify that we need to consider possible solutions, when $x\in [0, 8]$.
A: hint
The first thing to know in trigonometry
is
$$\cos^2 (something)+\sin^2 (something )=1$$
A: As $\csc^2A-\cot^2A=1$
$$3=4\csc(x+2)-3\cot^2(x+2)=4\csc(x+2)-3\{\csc^2(x+2)-1\}$$
$$\iff\csc(x+2)\{3\csc(x+2)-4\}=0$$
As for real $x+2,\csc(x+2)\ge1$ or $\le1$
$$3\csc(x+2)-4=0\iff\sin(x+2)=\dfrac34$$
A: HINT: since we have $$\sin(x+2)^2+\cos(x+2)^2=1$$ you have to solve
$$\sin(x+2)=\frac{3}{4}$$
