What is the general solution for solvefor x,e^{sin(x)}=e^{-sin(x)} ?

The general solution for solvefor x,e^{sin(x)}=e^{-sin(x)} is x=2pin,x=pi+2pin

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solvefor x,e^{sin(x)}=e^{-sin(x)}

solve for

x, e^{s i n (x)} = e^{- s i n (x)}

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

e^{s i n (x)} = e^{- s i n (x)}

Solve by substitution

e^{s i n (x)} = e^{- s i n (x)}

Let:

sin (x) = u

e^{u} = e^{- u}

e^{u} = e^{- u} : u = 0

e^{u} = e^{- u}

Apply exponent rules

e^{u} = e^{- u}

a^{f (x)} = a^{g (x)}

, then

f (x) = g (x)

u = - u

u = - u

Solve

u = - u : u = 0

u = - u

Move

u

to the left side

u = - u

Add

u

to both sides

u + u = - u + u

Simplify

2 u = 0

2 u = 0

Divide both sides by

2

2 u = 0

Divide both sides by

2

\frac{2 u}{2} = \frac{0}{2}

Simplify

u = 0

u = 0

u = 0

Substitute back

u = sin (x)

sin (x) = 0

sin (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

\frac{cos ( x )}{sin ( x )} = tan (x)

sin (B) = \frac{7 sin ( 3 6 ^{\circ} )}{12}

x, cos^{2} (x) = sin (y + \frac{π}{2})

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

sec^{2} (x) - tan^{2} (x) = cos (x)

What is the general solution for solvefor x,e^{sin(x)}=e^{-sin(x)} ?
The general solution for solvefor x,e^{sin(x)}=e^{-sin(x)} is x=2pin,x=pi+2pin