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

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

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sec(b/4)csc(b/4)=2csc(b/4)

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

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

Solution steps

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

Subtract

2 csc (\frac{b}{4})

from both sides

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

Factor

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

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

Solving each part separately

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

csc (\frac{b}{4}) = 0 :

No Solution

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

Combine all the solutions

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

2 cos (x) - 5 = 3 cos (x) - 4

\frac{π}{2} = arctan (x)

cos (n x) = 0

cos^{3} (θ) = cos (θ)

4 cos (2 x) = 22

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