What is the general solution for tan(2a)cot(a+20)=1 ?

The general solution for tan(2a)cot(a+20)=1 is a=360n+20,a=200+360n

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tan(2a)cot(a+20)=1

tan (2 a) cot (a + 2 0^{\circ}) = 1

a = 36 0^{\circ} n + 2 0^{\circ}, a = 20 0^{\circ} + 36 0^{\circ} n

Solution steps

tan (2 a) cot (a + 2 0^{\circ}) = 1

Subtract

1

from both sides

tan (2 a) cot (a + 2 0^{\circ}) - 1 = 0

Express with sin, cos

\frac{- cos ( 2 a ) sin ( \frac{18 0 ^{\circ} + 9 a}{9} ) + cos ( \frac{18 0 ^{\circ} + 9 a}{9} ) sin ( 2 a )}{cos ( 2 a ) sin ( \frac{18 0 ^{\circ} + 9 a}{9} )} = 0

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

- cos (2 a) sin (\frac{18 0 ^{\circ} + 9 a}{9}) + cos (\frac{18 0 ^{\circ} + 9 a}{9}) sin (2 a) = 0

Rewrite using trig identities

sin (2 a - \frac{18 0 ^{\circ} + 9 a}{9}) = 0

General solutions for

sin (2 a - \frac{18 0 ^{\circ} + 9 a}{9}) = 0

2 a - \frac{18 0 ^{\circ} + 9 a}{9} = 0 + 36 0^{\circ} n, 2 a - \frac{18 0 ^{\circ} + 9 a}{9} = 18 0^{\circ} + 36 0^{\circ} n

Solve

2 a - \frac{18 0 ^{\circ} + 9 a}{9} = 0 + 36 0^{\circ} n : a = 36 0^{\circ} n + 2 0^{\circ}

Solve

2 a - \frac{18 0 ^{\circ} + 9 a}{9} = 18 0^{\circ} + 36 0^{\circ} n : a = 20 0^{\circ} + 36 0^{\circ} n

a = 36 0^{\circ} n + 2 0^{\circ}, a = 20 0^{\circ} + 36 0^{\circ} n

2 sin (3 x) - 1 = 0, 0, 2 π

4 cos^{2} (x) = 2 cos (x) + 1

\frac{13}{sin ( 10 8 ^{\circ} )} = \frac{9}{sin ( x )}

tan^{2} (x) + 5 cos (x) - 8 = 0

3 sin (x) = 2 - 2 sin^{2} (x)

What is the general solution for tan(2a)cot(a+20)=1 ?
The general solution for tan(2a)cot(a+20)=1 is a=360n+20,a=200+360n