Let $f(x)=\sin^{23}x-\cos^{22}x$ and $g(x)=1+\frac{1}{2}\arctan|x|$,then the number of values of $x$ in the interval $[-10\pi,20\pi]$

Question

Let $f(x)=\sin^{23}x-\cos^{22}x$ and $g(x)=1+\frac{1}{2}\arctan|x|$,then the number of values of $x$ in the interval $[-10\pi,20\pi]$ satisfying the equation $f(x)=\text{sgn}(g(x)),$ is
$(A)6\hspace{1cm}(B)10\hspace{1cm}(C)15\hspace{1cm}(D)20$

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My Attempt:
$g(x)=1+\frac{1}{2}\arctan|x|$
$\text{sgn}(g(x))=1$
Because $g(x)$ is positive throughout the real number line.
$f(x)=\text{sgn}(g(x))\Rightarrow \sin^{23}x-\cos^{22}x=1$
We need to find the number of roots of $\sin^{23}x-\cos^{22}x-1=0$ in the interval $[-10\pi,20\pi]$
Periodicity of $\sin^{23}x-\cos^{22}x-1$ is $2\pi$.
But i am not sure if the function $\sin^{23}x-\cos^{22}x-1$ has one root in the interval $[0,2\pi]$ or not .
If it has one root in the $[0,2\pi]$ interval,then the answer must be $15$.
Am i correct or some other approach is applicable here. Please help me.

Answer 1

$$f(x)=sin^{23}x-cos^{22}x$$ then

$$f'(x)=sinxcosx\left(23sin^{21}x+22cos^{20}x\right)=sinxcosx \times g(x)$$

In First Quadrant obviously $f'(x) >0$, so $f$ is Increasing in $(0 \: \frac{\pi}{2})$

In Second Quadrant, $sinx \gt 0$ and $cosx \lt 0$ and $g(x) \gt 0$ so $f'(x) \lt 0$ in $(\frac{\pi}{2} \: \pi)$

and in Third and Fourth Quadrants, undoubtedly $f(x) \lt 0$