$\cos hx=\dfrac {e^x+e^{-x}}2 $ $$ \sin hx= \dfrac {e^x-e^{-x}}2$$

∴ Substituting it in given equations we get,

$ 7(\dfrac {e^x+e^{-x}}2)+8(\dfrac {e^x+e^{-x}}2)=1 \\ ∴7e^x+7e^{-x}+8e^x - 8e^{-x}=2 \\ ∴15e^x - e^{-x}=2$

Multiplying throughout by $e^x$ we get

$∴15e^{2x} - 1 = 2e^x \\ ∴15X^2-2X-1=0 $

Solving this quadratic equation we get

$X=\dfrac {2±\sqrt{4-4(15)(-1)}}{2(15) } \\ ∴X= \dfrac {1±8}{30} \\ ∴X= \dfrac 13 \space\space or \space\space \dfrac {-1}5 \\ ∴e^x= \dfrac 13 \space\space or \space\space \dfrac {-1}5 \\ ∴x=\log \dfrac 13 \space\space or\space\space x=\log(\dfrac {-1}{5}) $

Since x is real

$∴x=\log\dfrac 13$