Precalc: Determine the solutions of the equation cosx/1+sinx + 1+sinx/cosx=2 over interval [0,2pi)

Answer 1

skipping to step

2+2sin(x)=2cos(x)+2cos(x)sin(x)=cos(x)*(2+2sin(x))

dividing by (2+2sin(x)),,,,,where sin(x)≠-1,,,,,,,,,,, division by zero not allowed

wind up with cos(x)=1,,,,note when cos(x)=1, sin(x)=0

x=0 for cos(x)=1 in the domain interval specified.

Answer 2

cosx/(1 + sinx) + (1 + sinx)/cosx = 2; cos²x + 1 + 2sinx + sin²x = 2cosx + 2cosxsinx;

(1 - cosx)(1 + sinx) = 0; cosx = 1; x = 2πn, n = 0, ±1, ±2, ...

sinx = - 1, not in domain of this equation