A Level Maths Notes: FP2 – Solving Hyperbolic Trigonometric Equations (2)
An equation of the form
(1)
may have none, one or two solutions, unlike for the equivalent
ordinary trigonometric equation which may have many solutions.
We can solve equations of form (1) by substituting
and![]()
Then multiply though by
and
we have a quadratic equation in
which
we solve by the normal method of substitution and factorisation or
use of the quadratic formula.
Example: Solve the equation![]()
![]()
After simplification and
collection of like terms this becomes![]()
Multiply by
to
obtain
Substitute
to
obtain![]()
This expression factorises
to give![]()
We set each factor equal to
0 and solve for y, then use the original substitution to solve for![]()
![]()
![]()
Example: Solve the equation![]()
![]()
After simplification and
collection of like terms this becomes![]()
Multiply by
to
obtain
![]()
Substitute
to
obtain![]()
This expression does not
factorise and we must use the quadratic formula,![]()
![]()
![]()
to
3 sf.
which
has no value so the only solution is![]()