Best Answer:
f(x)=2cosx+sin2x

= 2* sin(x) -cos(2x) * d (2x)/dx = 2*sin(x) – 2*cos(2x) = 0

2(sin(x) – 2cos(2x) ) = 0

since cos(2x) = 1- sin^2(x) to make all term in sin(x)

2( sin(x) – 2(1 –2sin^2(x)) =0

4sin^2(x) + 2sin(x) -2 =0

2(2sin^2(x) +sin(x) -1 ) =

Use quadratic formula after factoring out the 2

Treat Sin(x) as normal x

A= 2, b=1 , c=-1

(-b +/- sqrt(b^2 -4ac) ) /2a

(-1 +/- sqrt( 1- 4*2* -1) /2(2)

(-1 +/- sqrt(9))/4

=(-1+/- 3)/4

= -1 and +1/2

x= arsin(-1) = -3π/2+ 2*π*k where k is an integer

and

arcsin (1/2) = π/6 + 2πk

and 5π/6 +2πk where k is an integer

a.

so the critical points

x= - 3π/2 + 2πk y= 2*cos(-3pi/2) + sin(6*pi) = 0

Critical Point series 1 ( - 3π/2 + 2πk, 0) where k is an integer

and

x= π/6 + 2πk y= 2*cos(pi/6)+ sin(2*pi/6) = 2*(sqrt(3)/2) + sqrt(3)/2 = 1.5*sqrt(3)

Critical Point series 2 ( π/6 + 2πk, 1.5*sqrt(3) ) where k is an integer

and

x = 5π/6 +2πk y =2*cos(5*pi)/6 +sin(2*pi/6) = -2*sqrt(3)/2 -sqrt(3)/2 = -1.5*sqrt(3)

Critical Point series 3

(5π/6 +2πk , -1.5*sqrt(3) ) where k is an integer

b.) Find the absolute min and max everywhere since no interval was specified.

From the graph and the values determined,

The absolute minimums are -1.5*sqrt(3) = approx. -2.598076211

The absolute minimums occur at 5π/6 +2πk where k is an integer

The absolute maximums are +1.5*sqrt(3) = approx. 2.598076211

The absolute maximums occur at π/6 + 2πk where k is an integer

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