Q:

Smallest value of $x$ satisfying $\sqrt {1+\sin 2x} =2\cos 3x $

Determine the smallest positive value of $x$ which satisfy the equation
$$\sqrt{1+\sin 2x} = 2\cos3x$$
What ever I got stuck.

A:

From
$\sqrt{1+\sin 2x}
=2\cos3x
$,
square both sides
(this may introduce some
extraneous solutions)
to get
$1+\sin 2x
=(2\cos3x)^2
=4\cos^23x
$.
Next,
use
$\sin 2x
=2\sin x \cos x
$
and
$\cos 3x 
=4\cos^3 x-3\cos x
$
to get
$\begin{array}\\
1+2\sin x \cos x
&=4(4\cos^3 x-3\cos x)^2\\
&=4\cos^2 x(4\cos^2 x-3)^2\\
\end{array}
$
At this point,
you can write
$2\sin x \cos x
=4\cos^2 x(4\cos^2 x-3)^2-1
$
and square this
to get
$4\sin^2 x \cos^2 x
=(4\cos^2 x(4\cos^2 x-3)^2-1)^2
$
or
$4(1-\cos^2 x) \cos^2 x
=(4\cos^2 x(4\cos^2 x-3)^2-1)^2
$.
Letting $y = \cos^2 x$,
this becomes
$4(1-y)y
=(4y(4y-3)^2-1)^2
$.
According to Wolfy,
this sextic has
real solutions
$y
= \frac12,
\frac38 \pm \frac{\sqrt{7}}{8}
$.
From $y = \frac12$,
$\cos x = \pm\sqrt{2}/2$,
so
$x = \pm\pi/4, \pm 3\pi/4$.
To see if this is a solution:
If $x = \pm\pi/4$,
$\sqrt{1+\sin 2x}
=\sqrt{1+\sin (\pm \pi/2)}
=0,\sqrt{2}
$
and
$2\cos(3x)
=2\cos(\pm 3\pi/4)
=-\sqrt{2}
$,
so no solution.
If $x = \pm 3\pi/4$,
$\sqrt{1+\sin 2x}
=\sqrt{1+\sin (\pm 3\pi/2)}
=0,\sqrt{2}
$
and
$2\cos(3x)
=2\cos(\pm 9\pi/4)
=2\cos(\pm \pi/4)
=\sqrt{2}
$,
so this has a solution.
I'll leave the other roots
to others.

