Find x for which 3cos2x + 4sin2x is minimum.?

i) Given one is as: f(x) = 3*cos(2x) + 4*sin(2x) = 5[(3/5)*cos(2x) + (4/5)*sin(2x)]
= 5[cos(2x)*cos(u) + sin(2x)*sin(u)] = 5*[cos(2x - u)], where u is given by tan⁻¹(4/3) = 53.13°]

ii) Max value of cos function is = 1, which is when the angle is = 0°
So here, 2x - u = 0°
2x = u° = 53.13°
==> x = 26.565°

Min value of cos function = -1, which is when the angle is = 180°
So x = (180 + u)/2 = 116.565°

Note: If you are satisfied with the solution kindly acknowledge; thanks for the same.

y = 3cos2x + 4sin2x = 5cos(2x - tan^-1(4/3))=5cos(2x-0.9273)
y min <---> 2x-0.9273 = pi +k 2pi or 2x-0.9273 = pi+k 2pi
solve for x

People here often fail to us correct symbology for the arguments of a function or for exponents. Here, it isn't clear whether you intend
.. 3*cos(2x) +4*sin(2x)
or
.. 3*cos^2(x) +4*sin^2(x) . . . . better written as 3*cos(x)^2 +4*sin(x)^2

Either way, a graphing calculator makes short work of this problem. On the interval [0, 360°], the answer to the first version is
.. {116.565°, 296.565°}
and the answer to the second is
.. {0°, 180°, 360°}

The second version is solved by inspection, as it reduces to
.. = 3 +sin(x)^2
which has a minimum where sin(x) = 0.

The first version is reduced by recognizing it is equivalent to
.. 5*sin(2x +arctan(3/4))
This has a minimum where
.. 2x +arctan(3/4) = 270° +n*360° . . . for some integer n
.. x = 135° -(1/2)*arctan(3/4) +n*180° . . . for some integer n
_____
On the graph at the source link, click on the curve to see its extreme points. Click on those to see their coordinates.

d[3*cos(2x) + 4*sin(2x)]/dx
= -6*sin(2x) + 8*cos(2x).
This is 0 when 3*sin(2x) = 4*cos(2x),
or when tan(2x) = 4/3, so when 2x
= 53.13 deg or 233.13 deg.
Only the latter is a minimum, the other is a maximum. So the "x" value must be
116.56 deg or 296.56 degrees, take that back to 2.03 radians or 5.18 radians.