What is the general solution for tan(3x)=cot(2x) ?

The general solution for tan(3x)=cot(2x) is x=(3pi}{10}+\frac{2pin)/5

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tan(3x)=cot(2x)

tan (3 x) = cot (2 x)

x = \frac{3 π}{10} + \frac{2 πn}{5}

Solution steps

tan (3 x) = cot (2 x)

Subtract

cot (2 x)

from both sides

tan (3 x) - cot (2 x) = 0

Express with sin, cos

\frac{- cos ( 2 x ) cos ( 3 x ) + sin ( 2 x ) sin ( 3 x )}{cos ( 3 x ) sin ( 2 x )} = 0

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

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

Rewrite using trig identities

- cos (2 x + 3 x) = 0

Divide both sides by

- 1

cos (2 x + 3 x) = 0

General solutions for

cos (2 x + 3 x) = 0

2 x + 3 x = \frac{π}{2} + 2 πn, 2 x + 3 x = \frac{3 π}{2} + 2 πn

Solve

2 x + 3 x = \frac{π}{2} + 2 πn : x = \frac{π}{10} + \frac{2 πn}{5}

Solve

2 x + 3 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{10} + \frac{2 πn}{5}

x = \frac{π}{10} + \frac{2 πn}{5}, x = \frac{3 π}{10} + \frac{2 πn}{5}

Since the equation is undefined for:

\frac{π}{10} + \frac{2 πn}{5}

x = \frac{3 π}{10} + \frac{2 πn}{5}

sin (z) = - 2

3 sin^{2} (x) + 4 cos (x) = 4

cos (θ) = \frac{1}{8}

2 tan^{2} (x) = 3 sec (x), 0 \leq x \leq 2 π

2 tan^{2} (x) + tan (x) = 0

What is the general solution for tan(3x)=cot(2x) ?
The general solution for tan(3x)=cot(2x) is x=(3pi}{10}+\frac{2pin)/5