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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Apply exponent rules

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

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

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

Apply log rule:

ln (e^{a}) = a

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

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

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

Solve

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

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

Move

1

to the right side

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

Add

1

to both sides

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

Simplify

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

Simplify

2 cos (x) - 1 + 1 : 2 cos (x)

2 cos (x) - 1 + 1

Add similar elements:

- 1 + 1 = 0

= 2 cos (x)

Simplify

ln (1) + 1 : 1

ln (1) + 1

Apply log rule:

lo g_{a} (1) = 0

= 0 + 1

Simplify

= 1

2 cos (x) = 1

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

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

Simplify

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

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

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

General solutions for

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

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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