What is the general solution for sec(β/4)csc(β/4)=2csc(β/4) ?

The general solution for sec(β/4)csc(β/4)=2csc(β/4) is β=(4pi)/3+8pin,β=(20pi)/3+8pin

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sec(β/4)csc(β/4)=2csc(β/4)

sec (\frac{β}{4}) csc (\frac{β}{4}) = 2 csc (\frac{β}{4})

β = \frac{4 π}{3} + 8 πn, β = \frac{20 π}{3} + 8 πn

Solution steps

sec (\frac{β}{4}) csc (\frac{β}{4}) = 2 csc (\frac{β}{4})

Subtract

2 csc (\frac{β}{4})

from both sides

sec (\frac{β}{4}) csc (\frac{β}{4}) - 2 csc (\frac{β}{4}) = 0

Factor

sec (\frac{β}{4}) csc (\frac{β}{4}) - 2 csc (\frac{β}{4}) : csc (\frac{β}{4}) (sec (\frac{β}{4}) - 2)

csc (\frac{β}{4}) (sec (\frac{β}{4}) - 2) = 0

Solving each part separately

csc (\frac{β}{4}) = 0 or sec (\frac{β}{4}) - 2 = 0

csc (\frac{β}{4}) = 0 :

No Solution

sec (\frac{β}{4}) - 2 = 0 : β = \frac{4 π}{3} + 8 πn, β = \frac{20 π}{3} + 8 πn

Combine all the solutions

β = \frac{4 π}{3} + 8 πn, β = \frac{20 π}{3} + 8 πn

sin (x) = \frac{1}{7}

tan (\frac{θ}{2}) = - 1

tan (\frac{θ}{2}) = 1

4 sin^{2} (x) cot (x) - 2 cot (x) = 0

cos (2 θ) + sin (θ) = 0

What is the general solution for sec(β/4)csc(β/4)=2csc(β/4) ?
The general solution for sec(β/4)csc(β/4)=2csc(β/4) is β=(4pi)/3+8pin,β=(20pi)/3+8pin