Finding the Maximum and Minimum of Trigonometric Equations By Quadratic Substitution y = 3 cos ⁡ 2 x + 7 sin ⁡ x + 2The original question was: Finding the stationary point of the function. I tried to use both differentiation and trigonometric substitution of cos(2x) into 1 − 2 sin 2 ⁡ ( x ).For the quadratic method, I substituted sin(x) as u and solved for u, where the the equation is equated to 0, and finding the midpoint of the roots to find out the value of x following the nature of a parabolic equation.Unfortunately I realized that trigonometric substitution eliminates a possible solution for x where cos ⁡ ( x ) = 0. I suspect that it might be due to a non function being converted into a function but I do not quite understand why does it causes so.

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Finding the Maximum and Minimum of Trigonometric Equations By Quadratic Substitution  y  =  3  cos  ⁡  2  x  +  7  sin  ⁡  x  +  2The original question was: Finding the stationary point of the function. I tried to use both differentiation and trigonometric substitution of cos(2x) into   1  −  2      sin    2    ⁡  (  x  ).For the quadratic method, I substituted sin(x) as u and solved for u, where the the equation is equated to 0, and finding the midpoint of the roots to find out the value of x following the nature of a parabolic equation.Unfortunately I realized that trigonometric substitution eliminates a possible solution for x where   cos  ⁡  (  x  )  =  0. I suspect that it might be due to a non function being converted into a function but I do not quite understand why does it causes so.

Kaeden Lara · Accepted Answer

Step 1First impulse is to just differentiate.  y  =  3  cos  ⁡  2  x  +  7  sin  ⁡  x  +  2 soStep 2                              y          ′                            =        −        6        sin        ⁡        (        2        x        )        +        7        cos        ⁡        x                                  =        −        12        sin        ⁡        (        x        )        cos        ⁡        (        x        )        +        7        cos        ⁡        x                                  =        cos        ⁡        (        x        )        (        −        12        sin        ⁡        (        x        )        +        7        )            so   cos  ⁡  (  x  )  =  0 or   sin  ⁡  (  x  )  =            7      12

klesstilne1 · Accepted Answer

Step 1With the substitution, you get the quadratic  5  +  7  u  −  6      u    2    ,for   −  1  ≤  u  ≤  1.. This has a maximum of 169/24, where   u  =  7      /    12  , and a minimum of −8, where   u  =  −  1.Step 2Hence if there are no points of inflection where the curve is stationary, then the stationary points occur where   sin  ⁡  x  =  −  1  , or where   sin  ⁡  x  =  7      /    12. One may then solve these for the possible values of x, if one so wishes.

Finding the Maximum and Minimum of Trigonometric Equations By Quadratic Substitution. y=3cos2x+7sin x+2

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