Answer:
would ensure that at least one real root exists for this equation when solving for .
Step-by-step explanation:
Apply the Pythagorean identity to replace the cosine this equation with sine:
.
Multiply both sides by to obtain:
.
.
If , then this equation would become a quadratic equation about :
.
However, for all real .
Hence, the value of must be between and (inclusive) for the original equation to have a real root when solving for .
Determinant of this quadratic equation about :
.
Hence, when solving for , the roots of in terms of would be:
.
.
Since , it is necessary that for the original solution to have a real root when solved for .
The first solution, , does not meet the requirements. On the other hand, simplifying , gives:
.
.
.
In other words, solving for would give a real root between if and only if .
On the other hand, given that for the in the original equation, solving that equation for would give a real root if and only if .
Therefore, the original equation with as the unknown has a real root if and only if .