Trigonometry

Question

Solve each of the following equations.

(i)   3cos² 2x+ 4 sin2x =1 , 0° < x < 360°

 

(ii)  sec x (tan x -2) = 2 cosec x , 0° < x < 360°

 

(iii) sin 3x(4 sin3x - 3 cos3x) =4, 0 ≤ x ≤ π (pi)

 

 

 

Ans: (i) 101.4°, 168.6°, 281.4°, 348.6°

       (ii) 69.9°, 143.8°, 249.9°, 323.8°

       (iii) π/6, π/2, 5π/6, 0.738, 1.79, 2.83

 

Thank you :)

 

Answer 1

(i)

3 cos^2(2x) + 4 sin(2x) = 1 , 0° < x < 360°

Let t = 2x, then 0° < t < 720°, and

3 cos^2(t) + 4 sin(t) = 1

Substituting cos^2(t) = 1 - sin^2(t):

3(1 - sin^2(t)) + 4 sin(t) = 1

3 - 3 sin^2(t) + 4 sin(t) - 1 = 0

3 sin^2(t) - 4 sin(t) - 2 = 0

Let u = sin(t)

3 u^2 - 4 u - 2 = 0

u = (4 +- sqrt(16+24))/(2(3))

u = (2 +- sqrt(4+6))/3

u = (2 +- sqrt(10))/3

sin(t) =? (2 +- sqrt(10))/3

(2 + sqrt(10))/3 = 1.72075922005613 > 1 so this can’t be sin(t)

(2 - sqrt(10))/3 = -0.38742588672279

Let s = arcsin((2-sqrt(10))/3) = -22.79442508271453°

Because 0° < t < 720°,

then t = 180°-s, 360°+s, 180°-s+360°, 360°+s+360°,

and x = t/2 = 90°-s/2, 180°+s/2, 270°-s/2, 360°+s/2

or x = 101.3972125°, 168.6027875°, 281.3972125°, 348.6027875°

 

Which agree with given answers of:

101.4°, 168.6°, 281.4°, 348.6°

(ii)

sec(x) (tan(x) - 2) = 2 csc(x) , 0° < x < 360°

divide both sides by csc(x):

tan(x) (tan(x) - 2) = 2

tan^2(x) - 2 tan(x) - 2 = 0

Let u = tan(x)

u^2 - 2u - 2 = 0

Complete the square:

u^2 + 2(-1)u + (-1)^2 - (-1)^2 - 2 = 0

(u-1)^2 = 3

|u-1| = sqrt(3)

u = tan(x) = 1 +- sqrt(3)

1 + sqrt(3) = 2.73205080756888

1 - sqrt(3) = -0.73205080756888

v1 = arctan(1 + sqrt(3)) = 69.89609063898293°

v2 = arctan(1 - sqrt(3)) = -36.20602311300322°

x = v1, v2+180°, v1+180°, v2+360°

x = 69.89609063898293°, 143.79397688699678°, 249.89609063898293°, 323.79397688699678°

 

Which agree with given answers of:

69.9°, 143.8°, 249.9°, 323.8°

(iii)

sin(3x) (4 sin(3x) - 3 cos(3x)) = 4, 0 ≤ x ≤ π (pi)

4 sin^2(3x) - 3 sin(3x)cos(3x) = 4 (sin^2(3x) + cos^2(3x))

- 3 sin(3x)cos(3x) = 4 cos^2(3x)

Divide both sides by cos^2(3x)

- 3 tan(3x) = 4

tan(3x) = -4/3

t = 3x so if 0 ≤ x ≤ π then 0 ≤ t ≤ 3π

u = tan^(-1)(-4/3) = -0.927295218001611

3x = t = pi + u, 2pi + u, 3pi + u

x = t/3 = (pi + u)/3, (2pi + u)/3, (3pi + u)/3

x = 0.73809914519606, 1.78529669639266, 2.83249424758926

Which agree with these given answers: 0.738, 1.79, 2.83

but not with these:
π/6 = 0.5235987755983
π/2 = 1.5707963267949
5π/6 = 2.6179938779915

Answer 2

i)

 3 ( cos 2X ) ^2  + 4 (Sin2x)  = 1

 

  3 ( 1 - Sin2X^2  )  + 4 Sin2X =1

 

  3 -3 Sin2X^ 2 + 4 Sin 2X - 1 = 0

 

    -3 Sin2X ^2 + 4 Sin 2X +2 =0

 

     3Sin2X ^2 - 4 Sin 2X - 2 = 0

 

        Sin 2X = 4/6 ±√( 16+24)/6=

 

        Sin 2X = 2/3 ± √10/3   

   

       Sin ^-1 ( 2/ 3 - √(10) / 3 ) = 337.2°       

 

         2X = 337.2 °    X =168.6°

         2 X = 180 + 22.8 = 202. 8   / 3rd Quadrant     X = 101.4°

 

 

    ii)   Sec X ( Tan X - 2 ) = 2 CosecX

 

      (1/ Cos X ) ( Sin X/ CosX - 2 ) = 2/ SinX

 

       Sin X  / CosX^2 - 2/ Cos X = 2 / Sin X

 

       SinX ^2 / CosX ^2 - 2SinX/ CosX - 2 =0

 

          Tan X ^2 - 2 tan X - 2 = 0

 

          Tan X = 1 ±√3

 

          tan X = 1 + √3     Tan ^-1 ( 1 +√3 ) = 69.9°, 180 + 69.9 = 249.9°

 

         Tan X = 1 - √3         Tan^-1( 1 - √3 ) = -36.2° or 323.8 °  , 180° - 36.2° = 143.8°

 

     iii )

 

         Sin3X ( 4 Sin3X - 3 Cos3X ) =4

 

           4 Sin3x^2 - 3 Sin 3X Cos 3X =4

           4 Sin3x^2 - 3/2 Sin 6X -4 =0

          This problem can be solved with inspection

 

        Maximum value of sin3x =1

 

        4 Sin X - 3Cos 3X , can have maximum value of

         4 (1) - 3 ( 0) =4

 

          So only possible solution for is

 

            Sin3 X = 1  Cos 3X =0

 

             Sin^-1 ( 3X=1 ) =  ∏/2 ,   X = ∏/6

 

             

 

       

      

              Rest later.