What is the general solution for tan(3x)*cot(x+40)=1 ?

The general solution for tan(3x)*cot(x+40)=1 is x=180n+20,x=110+180n

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tan(3x)*cot(x+40)=1

tan (3 x) \cdot cot (x + 4 0^{\circ}) = 1

x = 18 0^{\circ} n + 2 0^{\circ}, x = 11 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (3 x) cot (x + 4 0^{\circ}) = 1

Subtract

1

from both sides

tan (3 x) cot (x + 4 0^{\circ}) - 1 = 0

Express with sin, cos

\frac{- cos ( 3 x ) sin ( \frac{36 0 ^{\circ} + 9 x}{9} ) + cos ( \frac{36 0 ^{\circ} + 9 x}{9} ) sin ( 3 x )}{cos ( 3 x ) sin ( \frac{36 0 ^{\circ} + 9 x}{9} )} = 0

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

- cos (3 x) sin (\frac{36 0 ^{\circ} + 9 x}{9}) + cos (\frac{36 0 ^{\circ} + 9 x}{9}) sin (3 x) = 0

Rewrite using trig identities

sin (3 x - \frac{36 0 ^{\circ} + 9 x}{9}) = 0

General solutions for

sin (3 x - \frac{36 0 ^{\circ} + 9 x}{9}) = 0

3 x - \frac{36 0 ^{\circ} + 9 x}{9} = 0 + 36 0^{\circ} n, 3 x - \frac{36 0 ^{\circ} + 9 x}{9} = 18 0^{\circ} + 36 0^{\circ} n

Solve

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

Solve

3 x - \frac{36 0 ^{\circ} + 9 x}{9} = 18 0^{\circ} + 36 0^{\circ} n : x = 11 0^{\circ} + 18 0^{\circ} n

x = 18 0^{\circ} n + 2 0^{\circ}, x = 11 0^{\circ} + 18 0^{\circ} n

2 cos (θ) + 53 = 43

cos (x) = - \frac{1}{2} 2

sin (x) = - cos^{2} (x) - 1, 0 \leq x \leq 2 π

sin (x) (tan (x) - 3) = 0

sin (x) \cdot cos (x) = 2 cos^{2} (x)

What is the general solution for tan(3x)*cot(x+40)=1 ?
The general solution for tan(3x)*cot(x+40)=1 is x=180n+20,x=110+180n