# Solve equation 2^((sin x)^2)-2^((cosx)^2)=2?

### 2 Answers | Add Yours

You need to solve for x the given equation, hence, you should use the basic trigonometric formula to replace the exponent `cos^2 x` , such that:

`cos^2 x = 1 - sin^2 x`

`2^(sin^2 x) - 2^(1 - sin^2 x) = 2`

You should use exponentiation rules, such that:

`2^(1 - sin^2 x)= 2/(2^(sin^2 x))`

`2^(sin^2 x) - 2/(2^(sin^2 x)) = 2`

You should come up with the following substitution, such that:

`2^(sin^2 x) = t`

Replacing the variable yields:

`t - 2/t = 2 => t^2 - 2t - 2 = 0`

Using quadratic formula yields:

`t_(1,2) = (2+-sqrt(4 + 8))/2 => t_(1,2) = (2+-sqrt12)/2`

`t_(1,2) = (2+-2sqrt3)/2 => t_(1,2) = 1+-sqrt3`

You need to solve for x the equation `2^(sin^2 x) = 1+sqrt3` , such that:

`2^(sin^2 x) = 1+sqrt3`

Taking logarithms both sides yields:

`ln (2^(sin^2 x)) = ln(1+sqrt3) `

`sin^2 x*ln 2 = ln(1+sqrt3) `

`sin^2 x = ln(1+sqrt3)/ln 2 => sin^2 x ~~ 1.448 => sin x ~~ 1.2 ` invalid since `sin x <= 1` .

You cannot solve for x the equation `2^(sin^2 x) = 1-sqrt3` , since `1-sqrt3 < 0` and `2^(sin^2 x) > 0` , hence, the equation is invalid.

**Hence, evaluating the solutions to the given equation yields that the equation has no solutions.**

Solve `2^(sin^2x)-2^(cos^2x)=2` :

Note that `0<=sin^2x<=1` and `0<=cos^2x<=1`

`1<=2^(sin^2x)<=2` and `1<=2^(cos^2x)<=2`

The largest `2^(sin^2x)` can be is 2 while the smallest `2^(cos^2x)` can be is 1, so it is impossible that their difference is 2. The range of the difference is `-1<=2^(sin^2x)-2^(cos^2x)<=1`

**There can be no solution.**