What is the general solution for 2cos(2x)-4sin(x)+1=0 ?

The general solution for 2cos(2x)-4sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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2cos(2x)-4sin(x)+1=0

2 cos (2 x) - 4 sin (x) + 1 = 0

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

Solution steps

2 cos (2 x) - 4 sin (x) + 1 = 0

Rewrite using trig identities

3 - 4 sin (x) - 4 sin^{2} (x) = 0

Solve by substitution

sin (x) = - \frac{3}{2}, sin (x) = \frac{1}{2}

sin (x) = - \frac{3}{2} :

No Solution

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

Combine all the solutions

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

2 cos (x) - 2 sin (x) = 6

2 sin^{2} (θ) + 5 sin (θ) = 3

- 4 tan^{2} (θ) - 9 tan (θ) - 2 = 0

3 cos (θ) - 2 = 0

csc^{2} (\frac{x}{2}) + csc (\frac{x}{2}) = 4 + 4 sin (\frac{x}{2})

What is the general solution for 2cos(2x)-4sin(x)+1=0 ?
The general solution for 2cos(2x)-4sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin