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

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

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2sin(4θ)+1=0

2 sin (4 θ) + 1 = 0

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

Solution steps

2 sin (4 θ) + 1 = 0

Move

1

to the right side

2 sin (4 θ) = - 1

Divide both sides by

2

sin (4 θ) = - \frac{1}{2}

General solutions for

sin (4 θ) = - \frac{1}{2}

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

Solve

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

Solve

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

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

cot^{2} (x) csc^{2} (x) + 2 csc^{2} (x) - cot^{2} (x) = 2

cos (t) = \frac{3}{5}, 0 < t < \frac{π}{2}

cos (x) = \frac{- 2}{2}

y, - ln (cos (x)) = - 2 cos (x) \cdot sin (y)

2 sin (a) + csc (a) = 3

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