What is the general solution for 3cosh(2x)=5 ?

The general solution for 3cosh(2x)=5 is x= 1/2 ln(3),x=-1/2 ln(3)

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3cosh(2x)=5

3 cosh (2 x) = 5

x = \frac{1}{2} ln (3), x = - \frac{1}{2} ln (3)

Solution steps

3 cosh (2 x) = 5

Rewrite using trig identities

3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 5

3 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} = 5 : x = \frac{1}{2} ln (3), x = - \frac{1}{2} ln (3)

x = \frac{1}{2} ln (3), x = - \frac{1}{2} ln (3)

θ, 4 sin (θ) + 3 cos (θ) = 2

tan (x) n = 3

sin (x) = 0.6482078965

tan (2 θ) = \frac{2 tan ( θ )}{1 - tan ( θ )}

sec (θ) = 15, tan (2 θ) + 125 (cos (2 θ) - 1)

What is the general solution for 3cosh(2x)=5 ?
The general solution for 3cosh(2x)=5 is x= 1/2 ln(3),x=-1/2 ln(3)