Question 1196768
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Hi
 Q1. 2cos(π/4x)+√3 =0 
0r cos(π/4x) = -√3/2 
  | See wheel below (cosx, sinx)  
   cos(5π/6) = -√3/2   
      {{{5pi/6 =  pi/4x}}}    |Cross multiply and solve for x
      20πx = 6π
         x = 6/20 = 3/10

Q2. 4sin(π/4(x+2))+2 =0   0r sin (π/4(x+2)) = -2/4 = -1/2 
Look on Wheel:   sin (7π/6) = -1/2
{{{(pi/4(x+2)) = 7pi/6}}}  |Cross multiply and solve for x 
 *[illustration wheel].
Wish You the Best.
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