<s>[INST] Help Solving Trigonometry Equation I am having difficulties solving the following equation:
$$4\cos^2x=5-4\sin x$$
Hints on how to solve this equation would be helpful.
 [/INST] 
I am having difficulties solving the following equation:$${4\cos^2x=5-4\sin x}$$

First, substitute $4\cos^2(x)$ with $4\left(1 - \sin^2(x)\right)  = 4 - 4\sin^2(x).$ We are left with $4 - 4\sin^2(x)  = 5 - 4 \sin (x) .$ This can be rewritten as $-4\sin^2(x) + 4\sin(x) - 1 = 0.$
Observe that the equation can further be rewritten in the form $-4t^2 + 4t - 1 = 0$ where $t = \sin(x)$. Solve the quadratic equation for $t$ and then use $\sin(x) = t$ to solve for $x$.
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