What is the general solution for tan(8b)=cot(10b) ?

The general solution for tan(8b)=cot(10b) is b= pi/(36)+(pin)/9 ,b= pi/(12)+(pin)/9

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tan(8b)=cot(10b)

tan (8 b) = cot (10 b)

b = \frac{π}{36} + \frac{πn}{9}, b = \frac{π}{12} + \frac{πn}{9}

Solution steps

tan (8 b) = cot (10 b)

Subtract

cot (10 b)

from both sides

tan (8 b) - cot (10 b) = 0

Express with sin, cos

\frac{- cos ( 10 b ) cos ( 8 b ) + sin ( 10 b ) sin ( 8 b )}{cos ( 8 b ) sin ( 10 b )} = 0

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

- cos (10 b) cos (8 b) + sin (10 b) sin (8 b) = 0

Rewrite using trig identities

- cos (10 b + 8 b) = 0

Divide both sides by

- 1

cos (10 b + 8 b) = 0

General solutions for

cos (10 b + 8 b) = 0

10 b + 8 b = \frac{π}{2} + 2 πn, 10 b + 8 b = \frac{3 π}{2} + 2 πn

Solve

10 b + 8 b = \frac{π}{2} + 2 πn : b = \frac{π}{36} + \frac{πn}{9}

Solve

10 b + 8 b = \frac{3 π}{2} + 2 πn : b = \frac{π}{12} + \frac{πn}{9}

b = \frac{π}{36} + \frac{πn}{9}, b = \frac{π}{12} + \frac{πn}{9}

θ, cos (θ) = 0

sec^{2} (2 x) - 2 tan (2 x) = 0

x, d u = cos (4 x)

(\frac{sin ( θ )}{2 cos ( θ )}) - 1 = 0

2 \cdot cos^{2} (2 x) + 9 \cdot cos (4 x) = - 2.4214900000000 \dots

What is the general solution for tan(8b)=cot(10b) ?
The general solution for tan(8b)=cot(10b) is b= pi/(36)+(pin)/9 ,b= pi/(12)+(pin)/9