What is the general solution for tan(pi/4-x)+tan(x)-1=0 ?

The general solution for tan(pi/4-x)+tan(x)-1=0 is x=2pin,x=pi+2pin,x= pi/4+pin

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tan(pi/4-x)+tan(x)-1=0

tan (\frac{π}{4} - x) + tan (x) - 1 = 0

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

Solution steps

tan (\frac{π}{4} - x) + tan (x) - 1 = 0

Rewrite using trig identities

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

Simplify

\frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )} + tan (x) - 1 : \frac{- 2 sin ( x ) + tan ( x ) cos ( x ) + tan ( x ) sin ( x )}{cos ( x ) + sin ( x )}

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

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

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

Rewrite using trig identities

- sin (x) + sin (x) tan (x) = 0

Factor

- sin (x) + sin (x) tan (x) : sin (x) (tan (x) - 1)

sin (x) (tan (x) - 1) = 0

Solving each part separately

sin (x) = 0 or tan (x) - 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

Combine all the solutions

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

csc (\frac{2 θ}{3}) - 1 = 0

tan^{2} (θ) - 3 cot (θ) = 0

- 0.5 sin (t) = 0, 0 <, = t \leq 4 π

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

a, cos (a x) + (\frac{b}{x}) sin (a x) = 0

What is the general solution for tan(pi/4-x)+tan(x)-1=0 ?
The general solution for tan(pi/4-x)+tan(x)-1=0 is x=2pin,x=pi+2pin,x= pi/4+pin