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

The general solution for 0=1+2sin(2x) is x=(7pi)/(12)+pin,x=(11pi)/(12)+pin

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0=1+2sin(2x)

0 = 1 + 2 sin (2 x)

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn

Solution steps

0 = 1 + 2 sin (2 x)

Switch sides

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{7 π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

Solve

2 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{12} + πn

Solve

2 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{12} + πn

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn

0 = 1 + 2 cos (3 θ)

tan (4 x) = - 3

sin (π x) = - \frac{1}{2}

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

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

What is the general solution for 0=1+2sin(2x) ?
The general solution for 0=1+2sin(2x) is x=(7pi)/(12)+pin,x=(11pi)/(12)+pin