Determine the Relative Extrema of the Function $y = 2 \cos x + \sin 2 x$

question 48

Multiple Choice

Determine the relative extrema of the function $y = 2 \cos x + \sin 2 x$ on the interval $( 0,2 \pi )$ .

• A)relative minimum: \(\left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)\) relative maximum: \(\left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)\)
• B)relative minimum: \(\left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)\) relative maximum: \(\left( \frac { \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)\)
• C)relative minimum: \(\left( \frac { \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)\) relative maximum: \(\left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)\)
• D)relative minimum: \(\left( \frac { \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)\) relative maximum: \(\left( \frac { 5 \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)\)
• E)relative minimum: \(\left( \frac { 5 \pi } { 6 } , \frac { 3 \sqrt { 3 } } { 2 } \right)\) relative maximum: \(\left( \frac { \pi } { 6 } , - \frac { 3 \sqrt { 3 } } { 2 } \right)\)

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