What is the general solution for 1+cot(x)=csc(x) ?

The general solution for 1+cot(x)=csc(x) is x=2pin+pi/2

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1+cot(x)=csc(x)

1 + cot (x) = csc (x)

x = 2 πn + \frac{π}{2}

Solution steps

1 + cot (x) = csc (x)

Subtract

csc (x)

from both sides

1 + cot (x) - csc (x) = 0

Express with sin, cos

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

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

- 1 + cos (x) + sin (x) = 0

Rewrite using trig identities

- 1 + 2 sin (x + \frac{π}{4}) = 0

Move

1

to the right side

2 sin (x + \frac{π}{4}) = 1

Divide both sides by

2

sin (x + \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (x + \frac{π}{4}) = \frac{2}{2}

x + \frac{π}{4} = \frac{π}{4} + 2 πn, x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

Solve

x + \frac{π}{4} = \frac{π}{4} + 2 πn : x = 2 πn

Solve

x + \frac{π}{4} = \frac{3 π}{4} + 2 πn : x = 2 πn + \frac{π}{2}

x = 2 πn, x = 2 πn + \frac{π}{2}

Since the equation is undefined for:

2 πn

x = 2 πn + \frac{π}{2}

4 cos^{2} (x) - 42 cos (x) + 2 = 0

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

sin (θ) cos (θ) = cot (θ)

cot (θ) = - \frac{4}{3}

tan (x) = \frac{6}{10}

What is the general solution for 1+cot(x)=csc(x) ?
The general solution for 1+cot(x)=csc(x) is x=2pin+pi/2