$solve the equation \cos 2\theta =\sin \theta for o\le \theta \le 2\mathrm{\pi }$

$The equation of an inverse trig function drawn below is :$

$Wat is te domain and range of y={\cos }^{-1}\left(\frac{3x}{2}\right)?$

$$\begin{gathered} The diagram above shows the graph of: \\ \left(A\right) y={\sin }^{-1}\left(x+1\right) \left(B\right) y={\sin }^{-1}\left(x-1\right) \\ \left(C\right) y={\cos }^{-1}\left(x+1\right)-\frac{\mathrm{\pi }}{2} \left(D\right) y={\cos }^{-1}\left(x+1\right)-\frac{\mathrm{\pi }}{2} \end{gathered}$$

$Given y={\sin }^{-1}\left(\frac{2}{x}\right), which of the following is the expression for \frac{dy}{dx}?$

$A solution to the differential equation\frac{dy}{dx}=\frac{2}{\cos (x+y)+\cos (x-y)} can be obtained from:$

$$\begin{gathered} The exact value of {\sin }^{-1}\left(\sin \frac{4\mathrm{\pi }}{3}\right) is: \\ \left(A\right) \frac{4\mathrm{\pi }}{3} \left(B\right) \frac{\mathrm{\pi }}{3} \left(C\right) -\frac{\mathrm{\pi }}{3} \left(D\right) -\frac{4\mathrm{\pi }}{3} \end{gathered}$$

$The graph of y=ax and y={\tan }^{-1}\left(bx\right) intersect at three distinct point if$

$b) Solve \frac{dy}{dx}=\frac{y^2}{1+x^2} given y(0)=1$

$c) By letting t=\tan \frac{x}{2}, solve \sin x+\cos x=-1 in the domain 0\le x\le 2\mathrm{\pi }.$