Calculus help needed!

Using the formula, cos(2x)=1-2sin^2(x) and calculus, find the coordinates where the tangent to the curve of sin(2x)+2cos(x) is parallel to the x axis.

Take the first derivative of sin(2x)+2cos(x).

y' = 2cos2x - 2 sinx ... parallel to x-axis means slope = 0, so y' = 0

0 = 2cos2x - 2sinx ... divide out the 2 and add sinx to both sides

sinx = cos2x ... use the identity from the problem, put it on the right

sinx = 1-2sin^2(x) ... add/subtract right side to get 0 on the right and a quadratic

2sin^2(x) + sinx - 1 = 0 ... factor

(2sinx -1)(sinx + 1) = 0 ... set separately to zero

2sinx - 1 = 0, --> sinx = 1/2
sinx + 1 = 0, --> sinx = -1

Now solve for x. Your turn.

y=sin(2x)+2cos(x)
y' = 2cos(2x)-2sin(x) =0
2(1-2sin^2 x) - 2 sin(x) = 0
2-4sin^2(x)-2sin(x)=0
-4sin^2(x)-2sin(x)+2=0
4sin^2(x)+2sin(x)-2=0

4y^2+2y-2=0 ;letting y=sin(x)
2y^2+y-1=0
2y^2+2y-y-1=0
2y(y+1)-1(y+1)=0
(y+1)(2y-1)=0
y=1/2
y=-1

sin(x) = -1
x = 3π/2
sin(x) = 1/2
x= π/6, 5π/6

x= π/6, 5π/6, 3π/2
find y = sin(2x)+2cos(x) in each case .

dy / dx = 2 cos 2x - 2 sin x = 0 ---> { w = sin x } - 2 w² - w + 1 = 0...find w and then x

{ -π / 2 , π / 6 , 5π / 6 } + 2nπ