What is the general solution for 2sin(e^{t/4})+1=0 ?

The general solution for 2sin(e^{t/4})+1=0 is t=4ln((7pi)/6+2pin),t=4ln((11pi)/6+2pin)

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2sin(e^{t/4})+1=0

2 sin (e^{\frac{t}{4}}) + 1 = 0

t = 4 ln (\frac{7 π}{6} + 2 πn), t = 4 ln (\frac{11 π}{6} + 2 πn)

Solution steps

2 sin (e^{\frac{t}{4}}) + 1 = 0

Move

1

to the right side

2 sin (e^{\frac{t}{4}}) = - 1

Divide both sides by

2

sin (e^{\frac{t}{4}}) = - \frac{1}{2}

General solutions for

sin (e^{\frac{t}{4}}) = - \frac{1}{2}

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn, e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn

Solve

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn : t = 4 ln (\frac{7 π}{6} + 2 πn)

Solve

e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn : t = 4 ln (\frac{11 π}{6} + 2 πn)

t = 4 ln (\frac{7 π}{6} + 2 πn), t = 4 ln (\frac{11 π}{6} + 2 πn)

3 sin^{2} (x) - 2 sin (x) = 0

sin (u) = - \frac{2}{5}

2 sin (3 x) cos (2 x) - cos (2 x) = 0

\frac{1 + tan ( x )}{sin ( x )} = 0

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

What is the general solution for 2sin(e^{t/4})+1=0 ?
The general solution for 2sin(e^{t/4})+1=0 is t=4ln((7pi)/6+2pin),t=4ln((11pi)/6+2pin)