What is the general solution for 2sin(4x)+1=0 ?

The general solution for 2sin(4x)+1=0 is x=(7pi}{24}+(pin)/2 ,x=(11pi)/(24)+\frac{pin)/2

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2sin(4x)+1=0

2 sin (4 x) + 1 = 0

x = \frac{7 π}{24} + \frac{πn}{2}, x = \frac{11 π}{24} + \frac{πn}{2}

Solution steps

2 sin (4 x) + 1 = 0

Move

1

to the right side

2 sin (4 x) = - 1

Divide both sides by

2

sin (4 x) = - \frac{1}{2}

General solutions for

sin (4 x) = - \frac{1}{2}

4 x = \frac{7 π}{6} + 2 πn, 4 x = \frac{11 π}{6} + 2 πn

Solve

4 x = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{24} + \frac{πn}{2}

Solve

4 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{24} + \frac{πn}{2}

x = \frac{7 π}{24} + \frac{πn}{2}, x = \frac{11 π}{24} + \frac{πn}{2}

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

2 = 2 - 2 sin (θ)

sin (2 θ) - 3 cos (θ) = 0

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

sin (θ) - cos (2 θ) = 0

What is the general solution for 2sin(4x)+1=0 ?
The general solution for 2sin(4x)+1=0 is x=(7pi}{24}+(pin)/2 ,x=(11pi)/(24)+\frac{pin)/2