What is the general solution for tan(2x+1)=-cot(x+3) ?

The general solution for tan(2x+1)=-cot(x+3) is x=2pin+2+pi/2 ,x=2pin+2+(3pi)/2

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tan(2x+1)=-cot(x+3)

tan (2 x + 1) = - cot (x + 3)

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

Solution steps

tan (2 x + 1) = - cot (x + 3)

Subtract

- cot (x + 3)

from both sides

tan (2 x + 1) + cot (x + 3) = 0

Express with sin, cos

\frac{cos ( 1 + 2 x ) cos ( 3 + x ) + sin ( 1 + 2 x ) sin ( 3 + x )}{cos ( 1 + 2 x ) sin ( 3 + x )} = 0

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

cos (1 + 2 x) cos (3 + x) + sin (1 + 2 x) sin (3 + x) = 0

Rewrite using trig identities

cos (1 + 2 x - (3 + x)) = 0

General solutions for

cos (1 + 2 x - (3 + x)) = 0

1 + 2 x - (3 + x) = \frac{π}{2} + 2 πn, 1 + 2 x - (3 + x) = \frac{3 π}{2} + 2 πn

Solve

1 + 2 x - (3 + x) = \frac{π}{2} + 2 πn : x = 2 πn + 2 + \frac{π}{2}

Solve

1 + 2 x - (3 + x) = \frac{3 π}{2} + 2 πn : x = 2 πn + 2 + \frac{3 π}{2}

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

\frac{4 sec ^{2} ( x )}{2} = - 8 sec (x)

tan (x) + tan^{3} (x) = 0

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

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

(1 - sin (x)) \cdot cos^{3} (x) = 0

What is the general solution for tan(2x+1)=-cot(x+3) ?
The general solution for tan(2x+1)=-cot(x+3) is x=2pin+2+pi/2 ,x=2pin+2+(3pi)/2