Solving basic equations with cosh

Question

For example, $$3 \sinh(x)+4 \cosh(x)=4$$ I can either solve by rearranging for sinh $$(3 \sinh(x))^2=(4-4 \cosh(x))^2$$ (Let $c=\cosh(x)$) $$9(c^2-1)=16-c+16c^2$$ $$7c^2-32c+25=0$$ $$\cosh x = \frac{25}{7} \ or\ \cosh x=1$$ This gives solutions $x=0, x=\ln(\frac{25}{7}±\frac{24}{7})$ However, when solving for cosh instead initially (and working in terms of sinh), the solutions are $x=0, x=\ln(\frac{25}{7}+\frac{24}{7})$ only. When plugging to the original equation, the negative solution from the inverse cosh definition is the only solution that does not work.

Is it safe to assume that rearranging for cosh and solving in terms of sinh will provide the correct solutions, and that this is due to the cosh function not being one to one?

Answer 1

I think, it's better to make the following. $$\frac{3(e^x-e^{-x})}{2}+\frac{4(e^x+e^{-x})}{2}=4$$ or $$7e^{2x}-8e^x+1=0$$ or $$(7e^x-1)(e^x-1)=0,$$ which gives $$x=0$$ or $$x=-\ln7.$$

Answer 2

The additional root comes from squaring the equation $$3\sinh(x)=4(1-\cosh(x)) $$ The RHS is always non-positive since $\cosh(x)\geq 1$. On the other hand, LHS can be positive. If you want to obtain an equivalent equation after squaring, you need to impose the condition $\sinh(x)\leq 0\Leftrightarrow x\leq 0$. The principal values of the $\cosh^{-1}$ function are usually $[0,\infty)$. So $\cosh(x)=25/7$ gives $x=-\cosh^{-1}(25/7)=-\log(7)$ and we reject the root $\cosh^{-1}(25/7)$.