Solving Hyperbolic Equations

Solving a hyperbolic equation almost always comes down to one of two strategies: convert to a quadratic in $e^x$ex by substituting the exponential definitions, or convert to a quadratic in a single hyperbolic function by applying an identity. Both end with the same elementary algebra — a quadratic — and both demand the same care at the finish: rejecting any root with $e^x<0$ex<0, and giving the answer as an exact logarithm. This lesson drills both routes across the full range of question types, classifies how many solutions each equation has, and explains the $R\cosh(x+\alpha)$Rcosh(x+α) "harmonic form" as a stretch.

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1 Spec mapping

Solving hyperbolic equations is compulsory pure content for Papers 1 and 2 of AQA Further Mathematics (7367), and it is one of the most reliably examined skills in the whole strand. Carrying out the algebra is AO1; choosing the right reduction and reasoning about the number of solutions (and rejecting invalid roots) is AO2/AO3. It pulls together everything from the earlier lessons — the definitions, the identities, and the inverse logarithmic forms — and rests on solving quadratics and the logarithm laws from A-Level Maths. The recurring exam command is "solve, giving your answers in an exact form", and the marks are won by clean reduction, correct root rejection, and exact $\ln$ln answers. Because the topic synthesises the whole hyperbolic strand, it is a favourite for the longer, multi-step questions that discriminate between grades — a candidate who can fluently choose between the exponential route and the identity route, and who never forgets the rejection step, will pick up marks here that weaker candidates leave behind.

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2 Method 1 — reduce to a quadratic in $e^x$ex

The most direct route replaces each hyperbolic function by its exponential definition, multiplies through by $e^x$ex (or $e^{2x}$e2x) to clear negative powers, and recognises a quadratic in $u=e^x$u=ex. The decisive final step is rejecting any root $u\le0$u≤0, because $u=e^x>0$u=ex>0 for all real $x$x.

The reason this method always works is structural. Any hyperbolic function is a ratio of $e^x$ex and $e^{-x}$e−x, so once you clear the $e^{-x}$e−x terms (by multiplying through by an appropriate power of $e^x$ex), what remains is a polynomial in $e^x$ex — and for a single hyperbolic function it is a quadratic. The substitution $u=e^x$u=ex then turns the problem into ordinary GCSE/AS quadratic-solving, which you can do in your sleep; the only hyperbolic-specific subtlety is the positivity constraint $u>0$u>0. This is exactly why "convert to a quadratic in $e^x$ex" is the workhorse of the topic: it reduces every equation, however exotic-looking, to the same familiar endgame. The art lies only in choosing which power of $e^x$ex to multiply by (one power per negative-exponent term you need to clear) and in remembering the rejection step at the end.

Worked Example 1 — $\sinh x=k$sinhx=k

Solve $\sinh x=2$sinhx=2, giving the answer in exact form.

$\frac{e^x-e^{-x}}{2}=2\ \implies\ e^x-e^{-x}=4\ \implies\ e^{2x}-4e^x-1=0\quad(\times\,e^x).$2ex−e−x​=2 ⟹ ex−e−x=4 ⟹ e2x−4ex−1=0(×ex).

Let $u=e^x>0$u=ex>0: $u^2-4u-1=0$u2−4u−1=0, so $u=\dfrac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt5$u=24±16+4​​=2±5​. Since $2-\sqrt5<0$2−5​<0 is rejected, $e^x=2+\sqrt5$ex=2+5​, giving

$x=\ln\!\big(2+\sqrt5\big).$x=ln(2+5​).

Mark scheme: M1 substitute the definition; M1 form the quadratic in $e^x$ex; A1 $e^x=2+\sqrt5$ex=2+5​ (rejecting the negative root with reason); A1 $x=\ln(2+\sqrt5)$x=ln(2+5​). This single solution matches the fact that $\sinh$sinh is one-to-one.

Worked Example 2 — $\cosh x=k$coshx=k (two solutions)

Solve $\cosh x=3$coshx=3.

$\frac{e^x+e^{-x}}{2}=3\ \implies\ e^{2x}-6e^x+1=0\ \implies\ u=3\pm2\sqrt2.$2ex+e−x​=3 ⟹ e2x−6ex+1=0 ⟹ u=3±22​.

Both roots are positive ($3+2\sqrt2\approx5.83$3+22​≈5.83 and $3-2\sqrt2\approx0.17$3−22​≈0.17), so neither is rejected:

$e^x=3\pm2\sqrt2\ \implies\ x=\ln\!\big(3+2\sqrt2\big)\ \text{ or }\ x=\ln\!\big(3-2\sqrt2\big)=-\ln\!\big(3+2\sqrt2\big).$ex=3±22​ ⟹ x=ln(3+22​)  or  x=ln(3−22​)=−ln(3+22​).

So $x=\pm\ln(3+2\sqrt2)$x=±ln(3+22​), the two solutions expected because $\cosh$cosh is even.

Mark scheme: M1 quadratic in $e^x$ex; A1 $u=3\pm2\sqrt2$u=3±22​; A1 both $x=\pm\ln(3+2\sqrt2)$x=±ln(3+22​). (Note $\ln(3-2\sqrt2)=-\ln(3+2\sqrt2)$ln(3−22​)=−ln(3+22​) since the two roots are reciprocals — a tidy confirmation of the evenness.)

Worked Example 3 — $\tanh x=k$tanhx=k

Solve $\tanh x=\tfrac35$tanhx=53​.

$\frac{e^x-e^{-x}}{e^x+e^{-x}}=\tfrac35\ \implies\ 5(e^x-e^{-x})=3(e^x+e^{-x})\ \implies\ 2e^x=8e^{-x}\ \implies\ e^{2x}=4.$ex+e−xex−e−x​=53​ ⟹ 5(ex−e−x)=3(ex+e−x) ⟹ 2ex=8e−x ⟹ e2x=4.

$\therefore\ x=\tfrac12\ln 4=\ln 2.$∴ x=21​ln4=ln2.

Mark scheme: M1 substitute and cross-multiply; M1 collect to $e^{2x}=4$e2x=4; A1 $x=\ln 2$x=ln2. Since $|\tfrac35|<1$∣53​∣<1 there is exactly one solution, as expected for $\tanh$tanh.

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3 Method 2 — reduce to a quadratic in one hyperbolic function

When an equation mixes $\cosh^2$cosh2, $\sinh^2$sinh2, or $\operatorname{sech}$sech with $\tanh^2$tanh2, an identity collapses it to a quadratic in a single hyperbolic function, which is often quicker than going to exponentials. The key identities are $\cosh^2 x=1+\sinh^2 x$cosh2x=1+sinh2x and $\operatorname{sech}^2 x=1-\tanh^2 x$sech2x=1−tanh2x.

The crucial choice is which function to reduce to, and the rule is simple: reduce to whichever function appears to the first power. If the equation has $\cosh^2 x$cosh2x (an even power) and $\sinh x$sinhx (a first power), convert the $\cosh^2 x$cosh2x using $\cosh^2 x=1+\sinh^2 x$cosh2x=1+sinh2x so that everything is in $\sinh x$sinhx; the result is a quadratic in $\sinh x$sinhx. Conversely, if it has $\sinh^2 x$sinh2x and $\cosh x$coshx, convert the $\sinh^2 x=\cosh^2 x-1$sinh2x=cosh2x−1 to get a quadratic in $\cosh x$coshx. Picking the wrong direction leaves you with a mix of a function and its square root, which does not factorise — so this single decision is what makes the method work. Once you have a quadratic in one hyperbolic function, solve it, then convert each root back to $x$x using the appropriate inverse, remembering that a $\cosh x=k>1$coshx=k>1 root yields two values $\pm\operatorname{arcosh} k$±arcoshk whereas a $\sinh x=k$sinhx=k root yields just one.

Worked Example 4 — quadratic in $\sinh x$sinhx

Solve $2\cosh^2 x-5\sinh x=5$2cosh2x−5sinhx=5.

Substitute $\cosh^2 x=1+\sinh^2 x$cosh2x=1+sinh2x:

$2(1+\sinh^2 x)-5\sinh x=5\ \implies\ 2\sinh^2 x-5\sinh x-3=0\ \implies\ (2\sinh x+1)(\sinh x-3)=0.$2(1+sinh2x)−5sinhx=5 ⟹ 2sinh2x−5sinhx−3=0 ⟹ (2sinhx+1)(sinhx−3)=0.

So $\sinh x=-\tfrac12$sinhx=−21​ or $\sinh x=3$sinhx=3. Each gives a single $x$x (since $\sinh$sinh is one-to-one), via the inverse logarithmic form:

$\sinh x=-\tfrac12:\ x=\operatorname{arsinh}\!\big(-\tfrac12\big)=\ln\!\left(\frac{\sqrt5-1}{2}\right);\qquad \sinh x=3:\ x=\operatorname{arsinh}3=\ln\!\big(3+\sqrt{10}\big).$sinhx=−21​: x=arsinh(−21​)=ln(25​−1​);sinhx=3: x=arsinh3=ln(3+10​).

Mark scheme: M1 use $\cosh^2 x=1+\sinh^2 x$cosh2x=1+sinh2x; A1 quadratic $2\sinh^2 x-5\sinh x-3=0$2sinh2x−5sinhx−3=0; M1 factorise/solve; A1 both exact $x$x. (For $\sinh x=-\tfrac12$sinhx=−21​: $\operatorname{arsinh}(-\tfrac12)=\ln(-\tfrac12+\sqrt{\tfrac14+1})=\ln\tfrac{\sqrt5-1}{2}$arsinh(−21​)=ln(−21​+41​+1​)=ln25​−1​.)

Worked Example 5 — mixed $\cosh$cosh and $\sinh$sinh: go to exponentials

Solve $\cosh x-2\sinh x=3$coshx−2sinhx=3.

A mix of $\cosh$cosh and $\sinh$sinh (not their squares) is cleanest via exponentials:

$\frac{e^x+e^{-x}}{2}-2\cdot\frac{e^x-e^{-x}}{2}=3\ \implies\ \frac{-e^x+3e^{-x}}{2}=3\ \implies\ -e^x+3e^{-x}=6.$2ex+e−x​−2⋅2ex−e−x​=3 ⟹ 2−ex+3e−x​=3 ⟹ −ex+3e−x=6.

Multiply by $e^x$ex: $-e^{2x}+3=6e^x$−e2x+3=6ex, i.e. $e^{2x}+6e^x-3=0$e2x+6ex−3=0. Then $u=\dfrac{-6\pm\sqrt{36+12}}{2}=-3\pm2\sqrt3$u=2−6±36+12​​=−3±23​. Reject $-3-2\sqrt3<0$−3−23​<0; so $e^x=-3+2\sqrt3=2\sqrt3-3\ (\approx0.46>0)$ex=−3+23​=23​−3 (≈0.46>0):

$x=\ln\!\big(2\sqrt3-3\big).$x=ln(23​−3).

Mark scheme: M1 substitute definitions; M1 quadratic in $e^x$ex; A1 $e^x=2\sqrt3-3$ex=23​−3 (rejecting the negative root); A1 $x=\ln(2\sqrt3-3)$x=ln(23​−3).

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4 More worked examples

Worked Example 6 — quadratic with a zero root

Solve $\cosh^2 x-\sinh x=1$cosh2x−sinhx=1.

$(1+\sinh^2 x)-\sinh x=1\ \implies\ \sinh^2 x-\sinh x=0\ \implies\ \sinh x(\sinh x-1)=0.$(1+sinh2x)−sinhx=1 ⟹ sinh2x−sinhx=0 ⟹ sinhx(sinhx−1)=0.

So $\sinh x=0$sinhx=0 (giving $x=0$x=0) or $\sinh x=1$sinhx=1 (giving $x=\operatorname{arsinh}1=\ln(1+\sqrt2)$x=arsinh1=ln(1+2​)).

Mark scheme: M1 identity; A1 $\sinh x(\sinh x-1)=0$sinhx(sinhx−1)=0; A1 $x=0$x=0 and $x=\ln(1+\sqrt2)$x=ln(1+2​). Do not "divide by $\sinh x$sinhx" — that would lose the $x=0$x=0 solution, exactly the trap from polar coordinates.

Worked Example 7 — spot the recovery formula

Given $\cosh x+\sinh x=5$coshx+sinhx=5, find $x$x.

Recognise the recovery formula $\cosh x+\sinh x=e^x$coshx+sinhx=ex:

$e^x=5\ \implies\ x=\ln 5.$ex=5 ⟹ x=ln5.

Mark scheme: M1 recognise $\cosh x+\sinh x=e^x$coshx+sinhx=ex; A1 $x=\ln 5$x=ln5. Far slicker than substituting the full definitions — spotting the recovery formula saves a whole quadratic. The companion formula $\cosh x-\sinh x=e^{-x}$coshx−sinhx=e−x handles equations of the form $\cosh x-\sinh x=k$coshx−sinhx=k just as quickly: $e^{-x}=k$e−x=k gives $x=-\ln k$x=−lnk (valid for $k>0$k>0). Always scan a mixed $\cosh,\sinh$cosh,sinh equation for these patterns before reaching for the full exponential substitution.

Worked Example 8 — double-angle arguments

Solve $3\cosh 2x+5\sinh 2x=4$3cosh2x+5sinh2x=4.

Substitute the definitions at argument $2x$2x:

$3\cdot\frac{e^{2x}+e^{-2x}}{2}+5\cdot\frac{e^{2x}-e^{-2x}}{2}=4\ \implies\ \frac{8e^{2x}-2e^{-2x}}{2}=4\ \implies\ 4e^{2x}-e^{-2x}=4.$3⋅2e2x+e−2x​+5⋅2e2x−e−2x​=4 ⟹ 28e2x−2e−2x​=4 ⟹ 4e2x−e−2x=4.

Multiply by $e^{2x}$e2x: $4e^{4x}-4e^{2x}-1=0$4e4x−4e2x−1=0. Let $v=e^{2x}>0$v=e2x>0: $v=\dfrac{4\pm\sqrt{16+16}}{8}=\dfrac{4\pm4\sqrt2}{8}=\dfrac{1\pm\sqrt2}{2}$v=84±16+16​​=84±42​​=21±2​​. Reject $\dfrac{1-\sqrt2}{2}<0$21−2​​<0; so $e^{2x}=\dfrac{1+\sqrt2}{2}$e2x=21+2​​:

$x=\tfrac12\ln\!\left(\frac{1+\sqrt2}{2}\right).$x=21​ln(21+2​​).

Mark scheme: M1 definitions at $2x$2x; M1 quartic-in-$e^x$ex / quadratic in $v=e^{2x}$v=e2x; A1 $v=\tfrac{1+\sqrt2}{2}$v=21+2​​ (reject negative); A1 $x=\tfrac12\ln\tfrac{1+\sqrt2}{2}$x=21​ln21+2​​.

Worked Example 9 — a $\tanh$tanh/$\operatorname{sech}$sech equation

Solve $2\tanh^2 x+\operatorname{sech} x=1$2tanh2x+sechx=1.

Use $\tanh^2 x=1-\operatorname{sech}^2 x$tanh2x=1−sech2x:

$2(1-\operatorname{sech}^2 x)+\operatorname{sech} x=1\ \implies\ 2\operatorname{sech}^2 x-\operatorname{sech} x-1=0\ \implies\ (2\operatorname{sech} x+1)(\operatorname{sech} x-1)=0.$2(1−sech2x)+sechx=1 ⟹ 2sech2x−sechx−1=0 ⟹ (2sechx+1)(sechx−1)=0.

So $\operatorname{sech} x=-\tfrac12$sechx=−21​ (impossible, since $\operatorname{sech} x\in(0,1]$sechx∈(0,1]) or $\operatorname{sech} x=1$sechx=1 (so $\cosh x=1$coshx=1, $x=0$x=0). Hence $x=0$x=0.

Mark scheme: M1 use $\tanh^2 x=1-\operatorname{sech}^2 x$tanh2x=1−sech2x; A1 quadratic in $\operatorname{sech} x$sechx; M1 reject $\operatorname{sech} x=-\tfrac12$sechx=−21​ with reason; A1 $x=0$x=0. The range-based rejection of the negative root is a classic AO2 reasoning mark.

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5 How many solutions? A classification

Knowing the solution count in advance is a powerful check — and is itself examined. It follows directly from the graphs:

Equation	Number of real solutions
$\sinh x=k$sinhx=k	exactly 1 for every real $k$k ($\sinh$sinh is a bijection)
$\cosh x=k$coshx=k	0 if $k<1$k<1; 1 if $k=1$k=1; 2 if $k>1$k>1 (even, min value $1$1)
$\tanh x=k$tanhx=k	1 if $

So before solving $\cosh x=0.5$coshx=0.5 you should immediately state "no real solutions" (since $\cosh x\ge1$coshx≥1); attempting the quadratic would produce complex roots and waste time. Likewise $\tanh x=2$tanhx=2 has none. Stating the count first, then producing exactly that many roots, is the mark of a controlled answer.