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

The general solution for tan(β+10)=cot(2β-10) is β=30+(360n)/3 ,β=90+(360n)/3

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tan(β+10)=cot(2β-10)

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

β = 3 0^{\circ} + \frac{36 0 ^{\circ} n}{3}, β = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

Solution steps

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

Subtract

cot (2 β - 1 0^{\circ})

from both sides

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

Simplify

tan (β + 1 0^{\circ}) - cot (2 β - 1 0^{\circ}) : tan (\frac{18 β + 18 0 ^{\circ}}{18}) - cot (\frac{36 β - 18 0 ^{\circ}}{18})

tan (\frac{18 β + 18 0 ^{\circ}}{18}) - cot (\frac{36 β - 18 0 ^{\circ}}{18}) = 0

Express with sin, cos

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

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

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

Rewrite using trig identities

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

Divide both sides by

- 1

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

General solutions for

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

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

Solve

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

Solve

\frac{- 18 0 ^{\circ} + 36 β}{18} + \frac{18 0 ^{\circ} + 18 β}{18} = 27 0^{\circ} + 36 0^{\circ} n : β = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

β = 3 0^{\circ} + \frac{36 0 ^{\circ} n}{3}, β = 9 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

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

sin^{2} (x) = \frac{( 9 m - 4 )}{7}

cos (x) = \frac{3}{2}

6 sin^{2} (θ) - sin (θ) - 1 = 0

cos (x) = tan (x) sin (x)

What is the general solution for tan(β+10)=cot(2β-10) ?
The general solution for tan(β+10)=cot(2β-10) is β=30+(360n)/3 ,β=90+(360n)/3