Solve the following equations if 0<x<2pie (those are 'less than or equal to') answer to 2 decimal places.

a) square root of 2 cos(x = pie/2) + 1 = 0

b) sin(1/2)(x-2pie/9) = 0.6

a) sqrt 2cos(x-PI/2)=1 square both sides, then solve.
b) sin(x/2-PI/9)=.6
x/2= arc sin .6 + PI/9

a) To solve the equation √2cos(x-π/2) + 1 = 0, follow these steps:

1. Start with the equation: √2cos(x-π/2) + 1 = 0.
2. Subtract 1 from both sides of the equation to isolate the term with the cosine function: √2cos(x-π/2) = -1.
3. Square both sides of the equation to eliminate the square root: [√2cos(x-π/2)]² = (-1)².
This simplifies to 2cos²(x-π/2) = 1.
4. Divide both sides of the equation by 2 to isolate the cosine term: cos²(x-π/2) = 1/2.
5. Take the square root of both sides of the equation: cos(x-π/2) = ±√(1/2).
This gives two options: cos(x-π/2) = √(1/2) or cos(x-π/2) = -√(1/2).
6. Solve for x in each case:
- For cos(x-π/2) = √(1/2), find the corresponding angles within the given range (0<x<2π) where cos is equal to √(1/2). In this case, one solution is x = (π/2) + π/4 = 3π/4.
- For cos(x-π/2) = -√(1/2), find the corresponding angles within the given range (0<x<2π) where cos is equal to -√(1/2). In this case, one solution is x = (π/2) + 3π/4 = 5π/4.

Therefore, the solutions to the equation are x = 3π/4 and x = 5π/4.

b) To solve the equation sin((1/2)(x-2π/9)) = 0.6, follow these steps:

1. Start with the equation: sin((1/2)(x-2π/9)) = 0.6.
2. Divide both sides of the equation by 0.6 to isolate the sin function: (1/0.6)sin((1/2)(x-2π/9)) = 1.
3. Simplify the left side of the equation: sin((1/2)(x-2π/9)) = 10/6.
4. Take the arcsine of both sides of the equation to eliminate the sin function: (1/2)(x-2π/9) = arcsin(10/6).
5. Solve for x: x-2π/9 = 2arcsin(10/6).
6. Add 2π/9 to both sides of the equation: x = 2arcsin(10/6) + 2π/9.

Therefore, the solution to the equation is x = 2arcsin(10/6) + 2π/9, within the given range of 0<x<2π.