Solving Trigonometric Equation Problem im kinda stuck to solve the following problems below
Problem: 4cos²2x+sin2x=3   (0 < x <= π)
Steps:
4cos²2x+sin2x=3
2+2cos4x+sin2x=3
2cos4x+sin2x=1
2(1-2sin²2x)+sinx=1
2-4sin²2x+sinx=1
Bring every term to the right side
0= 1-2+4sin²2x-sinx
4sin²2x-sinx-1=0
Let M= sin2x, sin
4m²-m-1=0
X=+0.64 , x=-0.39

You have:
$$4\cos^2(2x)+\sin(2x) = 3$$
Use $\cos^2(2x) = 1-\sin^2(2x)$ then:
$$4(1-\sin^2(2x))+\sin(2x) = 3\implies4-4\sin^2(2x) + \sin(2x) = 3\implies\\1+\sin(2x)-4\sin^2(2x) = 0$$
which has roots:
$$\sin(2x) = \frac{1}{8}(1-\sqrt{17})\\\sin(2x) = \frac{1}{8}(1+\sqrt{17})$$
For the first equations:
$$\sin(2x) = \frac{1}{8}(1-\sqrt{17})\implies \sin^{-1}\sin(2x) = \sin^{-1}\frac{1}{8}(1-\sqrt{17})\implies \\2x = \sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)\implies x = \frac{\sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)}{2}$$
But there is also another solution in the other quadrant such that:
$$2x = \pi - \sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)\implies\\x = \frac{\pi}{2}-\frac{\sin^{-1}\left(\frac{1}{8}(1-\sqrt{17})\right)}{2}$$
You have to do the same thing for the second root:
$$\sin(2x) = \frac{1}{8}(1+\sqrt{17})$$
And then you have the answer
