What is the general solution for e^{2cos(x)-1}=1 ?

The general solution for e^{2cos(x)-1}=1 is x= pi/3+2pin,x=(5pi)/3+2pin

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e^{2cos(x)-1}=1

e^{2 c o s (x) - 1} = 1

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

e^{2 c o s (x) - 1} = 1

Apply exponent rules

2 cos (x) - 1 = ln (1)

Solve

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

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

General solutions for

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

csc (θ) = - \frac{37}{12}, \frac{3 π}{2} \leq θ \leq 2 π

\frac{3}{2} = 2 \cdot 25 sin (\frac{θ}{2})

sin^{2} (x) - sin (x) - 12 = 0

7 sin (2 ϕ) + 12 cos (ϕ) = 0

sec (\frac{θ}{2}) = cos (\frac{θ}{2})

What is the general solution for e^{2cos(x)-1}=1 ?
The general solution for e^{2cos(x)-1}=1 is x= pi/3+2pin,x=(5pi)/3+2pin