What is the general solution for solvefor a,P=cot^2(a) ?

The general solution for solvefor a,P=cot^2(a) is a=arccot(sqrt(P))+pin,a=arccot(-sqrt(P))+pin

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solvefor a,P=cot^2(a)

solve for

a, P = cot^{2} (a)

a = arccot (P) + πn, a = arccot (- P) + πn

Solution steps

P = cot^{2} (a)

Switch sides

cot^{2} (a) = P

Solve by substitution

cot^{2} (a) = P

Let:

cot (a) = u

u^{2} = P

u^{2} = P : u = P, u = - P

u^{2} = P

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = P, u = - P

Substitute back

u = cot (a)

cot (a) = P, cot (a) = - P

cot (a) = P, cot (a) = - P

cot (a) = P : a = arccot (P) + πn

cot (a) = P

Apply trig inverse properties

cot (a) = P

General solutions for

cot (a) = P

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot (P) + πn

a = arccot (P) + πn

cot (a) = - P : a = arccot (- P) + πn

cot (a) = - P

Apply trig inverse properties

cot (a) = - P

General solutions for

cot (a) = - P

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot (- P) + πn

a = arccot (- P) + πn

Combine all the solutions

a = arccot (P) + πn, a = arccot (- P) + πn

10 sin^{2} (2 u) + 6 cos^{2} (2 u) = 8

x, sin (x θ) = \frac{1}{2}

x, tan (x) = \frac{3.057}{6}

3 cos (4 5^{\circ}) + 4 cos (y) = 3

cos (3 θ) = cos (θ)

What is the general solution for solvefor a,P=cot^2(a) ?
The general solution for solvefor a,P=cot^2(a) is a=arccot(sqrt(P))+pin,a=arccot(-sqrt(P))+pin