Question

Express sinh 2x and Cosh 2x in terms of exponential function.​

Answer 1

Answer:

Step-by-step explanation:

2cosh2x−sinh2x=2(ex)2+2(e−x)22−(ex)2−(e−x)22

2=(ex)2+3(e−x)22

4=(ex)2+3(e−x)2

4=(ex)2+(3)(1)(ex)2

Multiplying both sides by (ex)2

4(ex)2=(ex)4+3

or

(ex)4−4(ex)2=−3

This sort of looks like something I could solve by completing the square or some other technique for solving a quadratic. But this is where I am stuck. I know that x=0 and x=0.549 are the solutions.