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I will give brainlest to right answer

There is an absolute maximum at x = π/2, absolute minimums at x = 7π/6 and 11π/6, and a relative maximum at x = 3π/2.

Step-by-step explanation:

f(x) = 2 sin x − cos(2x)

Relative extrema occur where f'(x) = 0.

f'(x) = 2 cos x − (-sin(2x) · 2)

f'(x) = 2 cos x + 2 sin(2x)

0 = 2 cos x + 2 sin(2x)

0 = cos x + sin(2x)

0 = cos x + 2 sin x cos x

0 = cos x (1 + 2 sin x)

cos x = 0 or sin x = -½

x = π/2, 7π/6, 3π/2, 11π/6

Evaluate the function at the endpoints and at each relative extrema.

f(0) = -1

f(π/2) = 3

f(7π/6) = -3/2

f(3π/2) = -1

f(11π/6) = -3/2

f(2π) = -1

There is an absolute maximum at x = π/2, absolute minimums at x = 7π/6 and 11π/6, and a relative maximum at x = 3π/2.

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step-by-step explanation: