Solve the equation 2coshx + sinhx = 2 given that x is a real number?

2coshx + sinhx = 2
2[(eˣ + e⁻ˣ)/2] + [(eˣ - e⁻ˣ)/2] = 2
2(eˣ + e⁻ˣ) + (eˣ - e⁻ˣ) = 4
2eˣ + 2e⁻ˣ + eˣ - e⁻ˣ = 4
3eˣ + e⁻ˣ = 4
[3eˣ + (1/eˣ)]eˣ = 4eˣ
3(eˣ)² + 1 = 4eˣ
3(eˣ)² - 4eˣ + 1 = 0
(3eˣ - 1)(eˣ - 1) = 0
eˣ = 1/3 or eˣ = 1
ln(eˣ) = ln(1/3) or eˣ = e⁰
xln(e) = ln(1/3) or x = 0
x = ln(1/3) or x = 0

coshx => (eˣ + e⁻ˣ)/2 and sinhx => (eˣ - e⁻ˣ)/2

so, 2coshx + sinhx => eˣ + e⁻ˣ + eˣ/2 - e⁻ˣ/2 = 3eˣ/2 + e⁻ˣ/2

i.e. 3eˣ/2 + e⁻ˣ/2 = 2

=> 3eˣ + e⁻ˣ = 4

Then, multiplying by eˣ we get:

3e²ˣ + 1 = 4eˣ

i.e. 3e²ˣ - 4eˣ + 1 = 0

=> (3eˣ - 1)(eˣ - 1) = 0

so, eˣ = 1/3 or 1

i.e. x = ln(1/3) or 0

:)>

2 ( e^x+e^(-x) /2) + (e^x - e^(-x)) /2 = 2
e^x + 1/e^x + e^x /2 -1/ (2e^x) = 2

Multiply both sides by 2e^x
2e^(2x) + 2 + e^(2x) - 1 = 4e^x
3e^(2x) - 4 e^x + 1 = 0

Let y=e^x
3y^2-4y+1 = 0
3y^2-3y-y+1 =0
3y(y-1) -1(y-1) = 0
(y-1) (3y-1) = 0
y=1
e^x = 1
x = 0
3y=1
y = 1/3
e^x = 1/3
x = ln(1/3) = -ln(3)

x = 0, -ln(3)