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Is there a method to find critical numbers of a trigonometric function ?

Posted on 2014-11-19
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Last Modified: 2014-11-21
f ( theta ) = 2 sec ( theta ) + tan ( theta )

0 <= theta < 2 pi

f ' ( theta ) = sec ( theta )  *   [ 2 tan ( theta ) + sec ( theta )  ]

set derivative to 0 to find critical numbers.

sec ( theta )  *   [ 2 tan ( theta ) + sec ( theta )  ] = 0

The only way I know how to do this problem is to plug different angle values for theta.  The values that
satisfy the equation are critical numbers.

Is there a better way to do this problem ?  Is there a method or procedure ?

This problem is from Advanced Placement High School Calculus class.
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Question by:naseeam
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by:ozo
ozo earned 325 total points
ID: 40454176
Since sec ( theta ) != 0,  you need [ 2 tan ( theta ) + sec ( theta )  ] = 0
which means 2 sin ( theta ) + 1 = 0
If you want to solve it numerically rather than analytically, and you plug different angle values for theta such that
2 sin ( theta1 ) + 1 > 0 and 2 sin ( theta2 ) + 1 < 0 then you know that 2 sin ( theta ) + 1 = 0 for some theta between theta1 and theta2, and you can binary search to narrow it down
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loneieagle2 earned 175 total points
ID: 40456435
2 sin (theta) + 1 = 0
2 sin (theta) = (-1)
sin(theta) = -1/2
theta= arcsin(-1/2)
with 0 <= theta <2pi, you should find two answers.
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