Assuming in tan(0.5) that 0.5 radians is intended, even though x is in degrees. If so, the solution for x is 83.3444° approx. This result was obtained using iterative methods. Other approximate solutions are:
27.3, 298.7, 316.4 degrees.
However, I think there is an error in the question: the equation should be:
tan(0.5x)+2sin(2x)=csc(x)
tan(x)=2tan(0.5x)/(1-tan2(0.5x)).
Let t=tan(0.5x), then tan(x)=2t/(1-t2), sin(x)=2t/(1+t2), cos(x)=(1-t2)/(1+t2),
sin(2x)=2sin(x)cos(x)=4t(1-t2)/(1+t2)2, csc(x)=1/sin(x)=(1+t2)/2t.
The equation becomes:
t+2(2sin(x)cos(x))=1/sin(x), t+8t(1-t2)/(1+t2)2=(1+t2)/2t,
8t(1-t2)/(1+t2)2=(1+t2)/2t-t=(1+t2-2t2)/2t=(1-t2)/2t,
(1-t2)[8t/(1+t2)2-1/2t)=0, t=±1 is a solution, that is, tan(x/2)=±1, x/2=±45°.
Therefore x=90° is a solution as is x=-90°, which is the same as 360-90=270°.
Check: tan(45)+2sin(180)=csc(90), 1+0=1, confirming x=90°,
tan(270)+2sin(540)=csc(270), -1+0=-1, confirming x=270°.
8t/(1+t2)2-1/2t=0 is also a solution, that is, 16t2=(1+t2)2,
±4t=1+t2, t2±4t+1=0, t=(±4±√12)/2=±2±√3=tan(x/2). tan(x)=2t/(1-t2).
So we have 4 values for t: -2-√3, -2+√3, 2-√3, 2+√3.
The corresponding values for x/2 are -75°, -15°, 15°, 75°, that is, x=
-150°, -30°, 30°, 150°, or 30°, 150°, 210°, 330°.
The complete solution is x=
30°, 90°, 150°, 210°, 270°, 330° in the range 0-360°.