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

The general solution for sin(x)(2cos(x)-1)=0 is x=2pin,x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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

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

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

Solution steps

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

2 cos (x) - 1 = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

cos (θ) = 1 + 2 sec (θ)

sin (3 θ) = - 1, 0 \leq θ < 2 π

cos (θ) = cos (arctan (\frac{3}{4}))

13 sin (2 x) - 3 = 3

sin (x) = \frac{4}{9}

What is the general solution for sin(x)(2cos(x)-1)=0 ?
The general solution for sin(x)(2cos(x)-1)=0 is x=2pin,x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin