sin2(x)-√cos(x)=0 or sin(2x)-√cos(x)=0?
Assume (1) sin2(x)-√cos(x)=0,
sin2(x)=√cos(x),
sin4(x)=cos(x),
(1-cos2(x))2=cos(x).
Let y=cos(x):
1-2y2+y4=y,
y4-2y2-y+1=0.
Newton's method:
Let f(y)=y4-2y2-y+1, f'(y)=4y3-4y-1.
yn+1=yn-f(yn)/f'(yn). Let y0=0.5:
y1=0.525, y2=0.5248..., y3=0.5248885..., ..., y=0.524888598656 approx.
cos(x)=0.524888598656, x=1.0182 radians approx. (58.34°)
Let y0=1.5: after several iterations y=1.4902161201, so, since cosine cannot exceed 1, we reject this solution.
x=1.0182 radians approx.
(2) sin(2x)-√cos(x)=0,
2sin(x)cos(x)=√cos(x),
4sin2(x)cos2(x)=cos(x).
So cos(x)=0 is one solution, and x=(2n+1)π/2 where n is an integer.
4sin2(x)cos(x)=1,
4(1-cos2(x))cos(x)=1, let y=cos(x):
4y-4y3=1, 4y3-4y+1=0.
From Newton's method, yn+1=yn-f(yn)/f'(yn). Let y0=0, then after several iterations:
cos(x)=y=0.269594436405, x=1.2978 radians (74.36°) approx.