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

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

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1+2sin(2x)=0

1 + 2 sin (2 x) = 0

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

2 sech^{2} (x) + tanh (x) = 0

8 cos (2 θ) = - 24 cos (θ) - 16

16 cos^{2} (θ) - 9 = 0

sin (a) + cos (a) = 1

2 sin^{2} (x) + 9 sin (x) - 5 = 0

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