Double root in Trigonometric equation?

Question

Consider the function $ f( x) = \sin^2 x + \sin x$ in the domain $ [0, 2 \pi]$, find the subset of the domain where $f \geq 0 $

Factoring by treating as quadratic in $ \sin$:

$$ f = ( \sin x + 1 ) ( \sin x)$$

Hence, the roots are $ \sin x = \{ 0, -1 \}$, hence the roots in $x$ are $ \{ 0, \pi , 2 \pi, \frac{3 \pi}{2} \}$. Let us partition the interval using the roots :

$$ \left[ 0, \pi \right] \cup \left[ \pi, \frac{ 3 \pi}{2} \right] \cup \left[ \frac{ 3 \pi}{2} , 2 \pi \right]$$

From regular algebra, we know that a 'root' corresponds to a sign flip, hence since the function is negative in the second interval of a domain partition, it must be positive in the third ... but actually it's not! i.e: $f \leq 0 $ for $ x \in [ \frac{ 3 \pi}{2} , 2 \pi ]$ but there is a problem, there was only a single root at $ x= \frac{ 3 \pi}{2}$.

So, how did the sign not flip here even when it crossed a root? That is what property of the trigonometric function is which causes this?

Answer 1

The property that causes this is the boundedness of the sine: there is only one solution per period of $\sin x=1$ and $\sin x=-1$, compared to two for $\sin x=a,a\in(-1,1)$.

If you plot the graphs $y=\sin x$ and $y=-1$ you will see that the latter line is tangent to the sine curve – i.e. $1+\sin x$ has only double root solutions.

Answer 2

Your factorization $ \ (\sin x)·(1 + \sin x) \ $ will also explain how the sign of $ \ f(x) \ $ behaves. Since $ \ (1 + \sin x) \ $ is non-negative for all values of $ \ x \ \ , $ the sign of $ \ f(x) \ $ is entirely determined by the sign of the factor $ \ (\sin x) \ \ . $ As this is negative in quadrants III and IV, $ \ f(x) \ $ is negative "to either side" of its zero at $ \ x \ = \ \frac{3 \pi}{2} \ \ . $

We can also consider the symmetry of the terms $ \ \sin^2 x \ + \ \sin x \ \ . $ Using the "angle-addition" formula for cosine, we note that $$ \cos \left(x \ - \ \frac{3 \pi}{2} \right) \ \ = \ \ \cos x · \cos \frac{3 \pi}{2} \ - \ \sin x · \sin \frac{3 \pi}{2} $$ $$ = \ \ \cos x · 0 \ - \ \sin x · (-1) \ \ = \ \ \sin x \ \ . $$

Each term in the expression for $ \ f(x) \ $ has "even symmetry" about the value $ \ x \ = \ \frac{3 \pi}{2} \ \ , $ so their sum also possesses this symmetry. Thus, there is no sign-change for $ \ f(x) \ $ on "passing through" $ \ x \ = \ \frac{3 \pi}{2} \ \ . $