Question

The number of solution of the equation 3^(sin2x+2cos2x)+ 3^(1-sin2x+2sin2x)=28 in [0,2π]?

The number of solution of the equation

#3^(sin2x+2cos2x)+ 3^(1-sin2x+2sin2x)=28# For `x in [0,2π]`

Answer 1

Zero

Explanation:

The maximum value of
`sin2x+2cos2x = sqrt5 sin(2x+tan^-1 2)`
is `sqrt5`, So, the maximum possible value of

`3^(sinx+2cos2x)`

is `3^sqrt5~~11.7`

On the other hand, the maximum possible value of `1-sin2x+2sin2x = 1+sin2x` is 2, so that the maximum possible value of this term is `9`

This means that the sum can not exceed `~~20.7`, let alone equal 28.