What is the general solution for tan((5pi)/4+x)+tan((5pi)/4-x)=4 ?

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

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tan((5pi)/4+x)+tan((5pi)/4-x)=4

tan (\frac{5 π}{4} + x) + tan (\frac{5 π}{4} - x) = 4

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

Solution steps

tan (\frac{5 π}{4} + x) + tan (\frac{5 π}{4} - x) = 4

Rewrite using trig identities

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

Subtract

4

from both sides

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

Simplify

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

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

- 8 cos (8 x) = 0

sec^{2} (x) - 4 = 0, 0 \leq x \leq 2 π

tan^{2} (x) + 7 tan (x) + 9 = 0

4 cos (x) = sec (x) + 3, 0 \leq x < 2 π

x, - \frac{1}{2 y ^{2}} = 3 sin (x) - \frac{1}{8}

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