What is the general solution for cos((5pi)/4+x)+sin((5pi)/4-x)=0 ?

The general solution for cos((5pi)/4+x)+sin((5pi)/4-x)=0 is x= pi/4+pin

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cos((5pi)/4+x)+sin((5pi)/4-x)=0

cos (\frac{5 π}{4} + x) + sin (\frac{5 π}{4} - x) = 0

x = \frac{π}{4} + πn

Solution steps

cos (\frac{5 π}{4} + x) + sin (\frac{5 π}{4} - x) = 0

Rewrite using trig identities

2 tan (x) - 2 = 0

Move

2

to the right side

2 tan (x) = 2

Divide both sides by

2

tan (x) = 1

General solutions for

tan (x) = 1

x = \frac{π}{4} + πn

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

θ, sin (θ) = \frac{1}{2}

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

6 cos (x) - 6 sin (x) = 0

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

What is the general solution for cos((5pi)/4+x)+sin((5pi)/4-x)=0 ?
The general solution for cos((5pi)/4+x)+sin((5pi)/4-x)=0 is x= pi/4+pin