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

The general solution for 1-2sin(2x)=0 is x= pi/(12)+pin,x=(5pi)/(12)+pin

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1-2sin(2x)=0

1 - 2 sin (2 x) = 0

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solution steps

1 - 2 sin (2 x) = 0

Move

1

to the right side

- 2 sin (2 x) = - 1

Divide both sides by

- 2

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

General solutions for

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

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

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

2 sin (x) + \frac{1}{sin ( x )} = 0

2 cos^{2} (θ) = 2 + sin (θ)

cos (x) - cos^{3} (x) = 0

cos (x) = - cos (x)

3 sin (x) sec (x) = 2 sin (x), 0 \leq x \leq 2 π

What is the general solution for 1-2sin(2x)=0 ?
The general solution for 1-2sin(2x)=0 is x= pi/(12)+pin,x=(5pi)/(12)+pin