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.
 A: Hint: You can use the identity: $$\cos^2(x) + \sin^2x =1.$$
Using substitution, you can then obtain a quadratic equation by letting $y = \sin x$.
A: 
$\cos^2(x) = 5-4 \sin(x)$

Move everything to the left hand side.

$\cos^2(x)-5+4 \sin(x) = 0$

Write in terms of $sin(x)$ using the identity $\cos^2(x) = 1-\sin^2(x)$:

$4 \sin(x)-4-\sin^2(x) = 0$

Factor constant terms from the left hand side and write the remainder as a square:

$-(\sin(x)-2)^2 = 0$

Multiply both sides by -1:

$(\sin(x)-2)^2 = 0$

Take the square root of both sides:

$\sin(x)-2 = 0$

Add 2 to both sides:
$\sin(x) = 2$ 
Your edited your post:

$$4 cos^2(x) = 5-4 sin(x)$$

Subtract $5-4 sin(x)$ from both sides:

$$4 cos^2(x)-5+4 sin(x) = 0$$

Using the identity $cos^2(x) = 1-sin^2(x):$

$$4 sin(x)-1-4 sin^2(x) = 0$$

Factor constant terms from the left hand side and write the remainder as a square:

$$-(2 sin(x)-1)^2 = 0$$

Multiply both sides by -1:

$$(2 sin(x)-1)^2 = 0$$

Take the square root of both sides:

$$2 sin(x)-1 = 0$$

Add 1 to both sides:

$$2 sin(x) = 1$$

Divide both sides by 2:

$$sin(x) = 1/2$$

A: Hint: Substitute in $\cos^2x=1-\sin^2x$, and solve the quadratic for $\sin x$.
A: 
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 $x$.
