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

The general solution for -4cos(2x)+4sin(x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

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

Rewrite using trig identities

- (1 - 2 sin^{2} (x)) \cdot 4 + 4 sin (x) = 0

Solve by substitution

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

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

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

6 sin (θ) = 1

- 5 sin (y) = 0

2 cos (\frac{3 x - π}{6}) + 1 = 0, 0 < x < 2 π

sin (α) = - 1

2 cos^{2} (a) + cos (a) - 1 = 0

What is the general solution for -4cos(2x)+4sin(x)=0 ?
The general solution for -4cos(2x)+4sin(x)=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin