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

The general solution for tan(pi/4-x)+cot(pi/4-x)=4 is x=(5pi)/6+pin,x= pi/6+pin

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tan(pi/4-x)+cot(pi/4-x)=4

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

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

Solution steps

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

Rewrite using trig identities

\frac{2 cos ^{2} ( x ) + 2 sin ^{2} ( x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) )} = 4

Subtract

4

from both sides

\frac{2 cos ^{2} ( x ) + 2 sin ^{2} ( x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) )} - 4 = 0

Simplify

\frac{2 cos ^{2} ( x ) + 2 sin ^{2} ( x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) )} - 4 : \frac{- 2 cos ^{2} ( x ) + 6 sin ^{2} ( x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) )}

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

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

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

Factor

- 2 cos^{2} (x) + 6 sin^{2} (x) : 2 (3 sin (x) + cos (x)) (3 sin (x) - cos (x))

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

Solving each part separately

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

3 sin (x) + cos (x) = 0 : x = \frac{5 π}{6} + πn

3 sin (x) - cos (x) = 0 : x = \frac{π}{6} + πn

Combine all the solutions

x = \frac{5 π}{6} + πn, x = \frac{π}{6} + πn

7 sin (θ) - 1 = 5 sin (θ)

csc (θ) = - \frac{7}{3}

sin (θ) = \frac{5}{8}

arctan (2 x) + arctan (x) = \frac{π}{4}

tan (a) = 1

What is the general solution for tan(pi/4-x)+cot(pi/4-x)=4 ?
The general solution for tan(pi/4-x)+cot(pi/4-x)=4 is x=(5pi)/6+pin,x= pi/6+pin