What is the general solution for sin(2x)+2sin(x)-cos(x)-1=0 ?

The general solution for sin(2x)+2sin(x)-cos(x)-1=0 is x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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sin(2x)+2sin(x)-cos(x)-1=0

sin (2 x) + 2 sin (x) - cos (x) - 1 = 0

x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

sin (2 x) + 2 sin (x) - cos (x) - 1 = 0

Rewrite using trig identities

- 1 - cos (x) + 2 sin (x) + 2 cos (x) sin (x) = 0

Factor

- 1 - cos (x) + 2 sin (x) + 2 cos (x) sin (x) : (cos (x) + 1) (2 sin (x) - 1)

(cos (x) + 1) (2 sin (x) - 1) = 0

Solving each part separately

cos (x) + 1 = 0 or 2 sin (x) - 1 = 0

cos (x) + 1 = 0 : x = π + 2 πn

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

cos (2 x) = - 0.4, - 18 0^{\circ} \leq x \leq 18 0^{\circ}

cos (x) = \frac{- 4}{\frac{8}{3} 3}

- 2 tan (x) + 6 = 8

2 sin (x) = 1.564

sin (x) = 0.08

What is the general solution for sin(2x)+2sin(x)-cos(x)-1=0 ?
The general solution for sin(2x)+2sin(x)-cos(x)-1=0 is x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin