What is the general solution for solvefor b,s(t)=tan(bt^2) ?

The general solution for solvefor b,s(t)=tan(bt^2) is b=(arctan(st))/(t^2)+(pin)/(t^2)

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solvefor b,s(t)=tan(bt^2)

solve for

b, s (t) = tan (b t^{2})

b = \frac{arctan ( s t )}{t ^{2}} + \frac{πn}{t ^{2}}

Solution steps

s (t) = tan (b t^{2})

Switch sides

tan (b t^{2}) = s t

Apply trig inverse properties

tan (b t^{2}) = s t

General solutions for

tan (b t^{2}) = s t

tan (x) = a \Rightarrow x = arctan (a) + πn

b t^{2} = arctan (s t) + πn

b t^{2} = arctan (s t) + πn

Solve

b t^{2} = arctan (s t) + πn : b = \frac{arctan ( s t )}{t ^{2}} + \frac{πn}{t ^{2}}; t \neq = 0

b t^{2} = arctan (s t) + πn

Divide both sides by

t^{2}; t \neq = 0

b t^{2} = arctan (s t) + πn

Divide both sides by

t^{2}; t \neq = 0

\frac{b t ^{2}}{t ^{2}} = \frac{arctan ( s t )}{t ^{2}} + \frac{πn}{t ^{2}}; t \neq = 0

Simplify

b = \frac{arctan ( s t )}{t ^{2}} + \frac{πn}{t ^{2}}; t \neq = 0

b = \frac{arctan ( s t )}{t ^{2}} + \frac{πn}{t ^{2}}; t \neq = 0

b = \frac{arctan ( s t )}{t ^{2}} + \frac{πn}{t ^{2}}

tan (x) = \frac{80}{90}

- \frac{5}{4} = cos (x)

sin (2 x) = cos^{2} (x)

4 cos^{2} (x) - 11 cos (x) + 6 = 0

tanh (x) = 0.99

What is the general solution for solvefor b,s(t)=tan(bt^2) ?
The general solution for solvefor b,s(t)=tan(bt^2) is b=(arctan(st))/(t^2)+(pin)/(t^2)