solvefor u,tan(u)+cot(u)= 1/f

Solution

solve for

u, tan (u) + cot (u) = \frac{1}{f}

u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn, u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

Solution steps

tan (u) + cot (u) = \frac{1}{f}

Subtract

\frac{1}{f}

from both sides

tan (u) + cot (u) - \frac{1}{f} = 0

Simplify

tan (u) + cot (u) - \frac{1}{f} : \frac{f tan ( u ) + f cot ( u ) - 1}{f}

\frac{f tan ( u ) + f cot ( u ) - 1}{f} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

f tan (u) + f cot (u) - 1 = 0

Rewrite using trig identities

- 1 + \frac{f}{cot ( u )} + cot (u) f = 0

Solve by substitution

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f}, cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f} : u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn

cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f} : u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

Combine all the solutions

u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn, u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

tan (3 x) + 3 = 0

sin (θ) = \frac{2 2}{3}

cos (θ) = \frac{1}{2}

cos (3 x) = sin (3 x)

sin (t) = \frac{3}{5}