What is the general solution for tan(2θ-10)=cot(θ) ?

The general solution for tan(2θ-10)=cot(θ) is θ=33.33333…+(360n)/3 ,θ=93.33333…+(360n)/3

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tan(2θ-10)=cot(θ)

tan (2 θ - 1 0^{\circ}) = cot (θ)

θ = 33.33333 \dots^{\circ} + \frac{36 0 ^{\circ} n}{3}, θ = 93.33333 \dots^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solution steps

tan (2 θ - 1 0^{\circ}) = cot (θ)

Subtract

cot (θ)

from both sides

tan (2 θ - 1 0^{\circ}) - cot (θ) = 0

Express with sin, cos

\frac{- cos ( \frac{- 18 0 ^{\circ} + 36 θ}{18} ) cos ( θ ) + sin ( \frac{- 18 0 ^{\circ} + 36 θ}{18} ) sin ( θ )}{cos ( \frac{- 18 0 ^{\circ} + 36 θ}{18} ) sin ( θ )} = 0

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

- cos (\frac{- 18 0 ^{\circ} + 36 θ}{18}) cos (θ) + sin (\frac{- 18 0 ^{\circ} + 36 θ}{18}) sin (θ) = 0

Rewrite using trig identities

- cos (\frac{- 18 0 ^{\circ} + 36 θ}{18} + θ) = 0

Divide both sides by

- 1

cos (\frac{- 18 0 ^{\circ} + 36 θ}{18} + θ) = 0

General solutions for

cos (\frac{- 18 0 ^{\circ} + 36 θ}{18} + θ) = 0

\frac{- 18 0 ^{\circ} + 36 θ}{18} + θ = 9 0^{\circ} + 36 0^{\circ} n, \frac{- 18 0 ^{\circ} + 36 θ}{18} + θ = 27 0^{\circ} + 36 0^{\circ} n

Solve

\frac{- 18 0 ^{\circ} + 36 θ}{18} + θ = 9 0^{\circ} + 36 0^{\circ} n : θ = 33.33333 \dots^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solve

\frac{- 18 0 ^{\circ} + 36 θ}{18} + θ = 27 0^{\circ} + 36 0^{\circ} n : θ = 93.33333 \dots^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 33.33333 \dots^{\circ} + \frac{36 0 ^{\circ} n}{3}, θ = 93.33333 \dots^{\circ} + \frac{36 0 ^{\circ} n}{3}

sinh (x) = 4

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, 0 \leq θ < 2 π

cos (x) sin (x) = 1

24 arctan (x) = 4 π

cos^{2} (x) - sin (x) - 1 = 0

What is the general solution for tan(2θ-10)=cot(θ) ?
The general solution for tan(2θ-10)=cot(θ) is θ=33.33333…+(360n)/3 ,θ=93.33333…+(360n)/3