<s>[INST] 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.
 [/INST] 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 $\le1$
$$3\csc(x+2)-4=0\iff\sin(x+2)=\dfrac34$$
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