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

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

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2sin(2x)+1=0

2 sin (2 x) + 1 = 0

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

Solution steps

2 sin (2 x) + 1 = 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

sin (θ) = \frac{5}{13}

sin^{2} (x) = 2 sin (x) + 3

2 cos^{2} (θ) - sin (θ) = 1

cosh (x) = 0

cos (x) = \frac{5}{13}

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