Question #a2fd4

1 Answer
Cesareo R.
Dec 12, 2017

See below.

Explanation:

#2^x = e^(lambda x)# then

#e^sinx = 2cosh(lambda x)# or

#sinx = log_e(2cosh(lambda x))#

Here #lambda = log_e 2#

This equation does not have real solutions as can be observed in the following plot

enter image source here

In blue # log_e(2cosh(lambda x))# and in red #sinx#

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