Solve the equation for exact solutions in the interval [0,360). sin2Theta = 2cos^2Theta

sin^2 (theta)  = 2cos^2(theta)

sin^2(theta)  = 2[ 1 - sin^2(theta)]

sin^2(theta)  = 2 - 2 sin^2(theta)       add 2 sin^2(theta)  to both sides

3sin^2(theta)   = 2     divide both sides by 3

sin^2(theta)  = 2/3      take the positive/negative sqrt 

sin(theta)  =  sqrt (2/3)          and       sin(theta)  =  - sqrt(2/3)

Using the arcsin  we have

arcsin[ sqrt(2/3)]   = about 54.74°   and  about  (180 - 54.74)°  = 125.26°

arcsin [ - sqrt(2/3)]  =  about  - 54.74°   =  (180 + 54.74) ° =  about 234.74°   and about (360 - 54.74)° = about 305.26°

Here's a graph of the intersection points of the two functions........https://www.desmos.com/calculator/8bcyqeldft

[ For your second question, what is "arcc"  ???]

Sorry i ment to say 2arccos not arcc

and thank you very much!

Here's the second one :

 arcsinx-2arccos sqrt(3)/2 = pi/3           [ arccos sqrt(3/2)  has two values on [0, 2pi] .......pi/6    and 11pi/6 ]

So we have

arc sinx - 2[pi6]   = pi/3

arcsinx - pi/3  =   pi/3       add pi/3 to both sides

arcsinx =   2pi/3

Notice that no solution exists, because the sine's greatest value =  1....so we are trying to find an angle whose sine = 2pi/3 .......  and  2pi/3 > 1, so this doesn't angle exist

Also........if we calculated this for 11pi/ 6, we would have the same problem

arcsinx - 2 [11pi/6]  = pi/3

arcsinx - 11pi/3 =  pi/3

arcsinx = 12pi/3    = 4pi       and this is much greater than 1

So........no solution exists for the second problem