\begin{tabular}{l} (b) Using the formula \( \operatorname{Sin} 2 A=2 \operatorname{Sin} A \operatorname{Cos} A \) \\ solve the equation \( \sin 2 x-\cos x=0 \) \\ in the interval \( 0^{\circ} \leq x \leq 360^{\circ} \) \\ \hline\end{tabular}

\( \frac { 4 \sin \frac { \alpha } { 2 } \cos \frac { \alpha } { 2 } } { \tan \frac { \alpha } { 2 } } - 1 \)

If \( \theta=\frac{5 \pi}{6} \), then \( \cos (\theta)=\square \) \( \sin (\theta)=\square \) Give exact values. No decimals allowed!

Find an angle \( \theta \) coterminal to \( -555^{\circ} \), where \( 0^{\circ} \leq \theta<360^{\circ} \)

If \( \cos (\alpha+\beta)=0 \) then \( \sin (\alpha+2 \beta) \)

Fid an angle \( \theta \) coterminal to \( -561^{\circ} \), where \( 0^{\circ} \leq \theta<360^{\circ} \)