F.4 Maths Trigo

2 cos x = sin x +1
(2 cos x)^2 = (sin x +1)^2
4 (cos x)^2 = (sin x)^2+ 2 sin x + 1
4 [1- (sin x)^2] = (sin x)^2+ 2 sin x + 1
4- 4(sin x)^2 = (sin x)^2+ 2 sin x + 1
5(sin x)^2 + 2 sin x - 3 = 0
(5sin x -3) (sin x +1) = 0
sin x = 3/5 or -1
x= 36.870 or 143.13 or 270

2cos x = sinx +1
(2cos x)^2 = (sinx +1)^2
4(cos x)^2 = (sinx)^2+ 2sin x + 1
4[1- (sinx)^2] = (sinx)^2+ 2sin x + 1
4- 4(sin x)^2 = (sin x)^2+ 2 sin x + 1
5(sin x)^2 + 2 sin x - 3 = 0
(5sin x -3) (sin x +1) = 0
sin x = 3/5 or sin x = -1 (rejected as 0＜x＜180)
x= 36.9 or 143.1 (rejected since 2cos x = sinx +1 → cosx = (sinx+1)/2 ＞＝(-1+1)/2＞＝0 )
therefore x = 36.9

(＞＝ means greater than or equal to)

if you cant understand why we should reject 143.1
you may sub it back to the equation and you will find that LHS ≠ RHS

2cosx=sinx+1
2cosx-sinx=1
√5[cosx(2/√5)-sinx(1/√5)]=1
√5cos(x+26.6˚)=1
cos(x+26.6˚)=1/√5
x+26.6˚=63.4˚
x=36.8˚