Question

solve the equation
2cos^2x=1+sinx for 0<_ x<_2r(pi)

Answer 1

us cos*2 x + sin*2 x =1 from this cos*2 x =1-sin*2 x hence your equation will be
2(1-sin*2 x)=1+sin x --- 2-2sin*2 x=1+sin x --- 2sin*2 x +sin x -1 =0
sign sin x with a and you get 2a*2 +a-1=0 --- a(2a+1)=1 --- a=1
2a+1=1 --- 2a=0 --- a=0
a=1 --- sin x=1 and sin x=0