2cos^2x=1+sinx

Question

solve for 0 less than or equal to x less than or equal to 2pi.  2cos^x=1+sinx

Answer 1

2cos²(x)=2(1-sin²(x));

2(1-sin²(x))=1+sin(x),

2-2sin²(x)=1+sin(x),

2sin²(x)+sin(x)-1=0=(2sin(x)-1)(sin(x)+1),

sin(x)=½, sin(x)=-1,

x=π/6, 5π/6, 3π/2.