What is the general solution for solvefor g,θ(t)=-1cos(sqrt(g/l)t) ?

The general solution for solvefor g,θ(t)=-1cos(sqrt(g/l)t) is

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solvefor g,θ(t)=-1cos(sqrt(g/l)t)

solve for

g, θ (t) = - 1 cos (\frac{g}{l} t)

g = \frac{l arccos ^{2} ( - θt )}{t ^{2}} + \frac{4 π l n arccos ( - θt )}{t ^{2}} + \frac{4 π ^{2} l n ^{2}}{t ^{2}}, g = \frac{l arccos ^{2} ( - θt )}{t ^{2}} - \frac{4 π l n arccos ( - θt )}{t ^{2}} + \frac{4 π ^{2} l n ^{2}}{t ^{2}}

Solution steps

θ (t) = - 1 \cdot cos (\frac{g}{l} t)

Switch sides

- 1 \cdot cos (\frac{g}{l} t) = θt

Divide both sides by

- 1

cos (\frac{g}{l} t) = - θt

Apply trig inverse properties

\frac{g}{l} t = arccos (- θt) + 2 πn, \frac{g}{l} t = - arccos (- θt) + 2 πn

Solve

\frac{g}{l} t = arccos (- θt) + 2 πn : g = \frac{l arccos ^{2} ( - θt )}{t ^{2}} + \frac{4 π l n arccos ( - θt )}{t ^{2}} + \frac{4 π ^{2} l n ^{2}}{t ^{2}}

Solve

\frac{g}{l} t = - arccos (- θt) + 2 πn : g = \frac{l arccos ^{2} ( - θt )}{t ^{2}} - \frac{4 π l n arccos ( - θt )}{t ^{2}} + \frac{4 π ^{2} l n ^{2}}{t ^{2}}

g = \frac{l arccos ^{2} ( - θt )}{t ^{2}} + \frac{4 π l n arccos ( - θt )}{t ^{2}} + \frac{4 π ^{2} l n ^{2}}{t ^{2}}, g = \frac{l arccos ^{2} ( - θt )}{t ^{2}} - \frac{4 π l n arccos ( - θt )}{t ^{2}} + \frac{4 π ^{2} l n ^{2}}{t ^{2}}

sec (x) = - 3

2 tan (x) csc (x) + 2 csc (x) + tan (x) + 1 = 0

csc (x) + cot (x) = 3, 0 \leq x \leq 2 π

(tan^{2} (x) - 4) (2 cos (x) + 1) = 0

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

What is the general solution for solvefor g,θ(t)=-1cos(sqrt(g/l)t) ?
The general solution for solvefor g,θ(t)=-1cos(sqrt(g/l)t) is