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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Popular Trigonometry >

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

Solution

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

Solution

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

Degrees

t = 0^{\circ} + 526.52744 \dots^{\circ} n, t = 0^{\circ} + 570.31408 \dots^{\circ} n

Solution steps

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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)

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

Multiply both sides by

6

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

Simplify

e^{\frac{t}{4}} \cdot 6 = 7 π + 12 πn

Divide both sides by

6

e^{\frac{t}{4}} \cdot 6 = 7 π + 12 πn

Divide both sides by

6

\frac{e ^{\frac{t}{4}} \cdot 6}{6} = \frac{7 π}{6} + \frac{12 πn}{6}

Simplify

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

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

Apply exponent rules

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

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

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

Apply log rule:

ln (e^{a}) = a

ln (e^{\frac{t}{4}}) = \frac{t}{4}

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

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

Solve

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

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

Multiply both sides by

4

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

Multiply both sides by

4

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

Simplify

t = 4 ln (\frac{7 π}{6} + 2 πn)

t = 4 ln (\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)

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

Multiply both sides by

6

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

Simplify

e^{\frac{t}{4}} \cdot 6 = 11 π + 12 πn

Divide both sides by

6

e^{\frac{t}{4}} \cdot 6 = 11 π + 12 πn

Divide both sides by

6

\frac{e ^{\frac{t}{4}} \cdot 6}{6} = \frac{11 π}{6} + \frac{12 πn}{6}

Simplify

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

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

Apply exponent rules

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

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

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

Apply log rule:

ln (e^{a}) = a

ln (e^{\frac{t}{4}}) = \frac{t}{4}

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

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

Solve

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

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

Multiply both sides by

4

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

Multiply both sides by

4

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

Simplify

t = 4 ln (\frac{11 π}{6} + 2 πn)

t = 4 ln (\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)

Graph

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Frequently Asked Questions (FAQ)

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)