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Mathematics Dictionary
Dr. K. G. Shih
Trigonometric Equations
Subjects


  • TR 09 00 | - Outlines
  • TR 09 01 | - Solve sin(x) = a
  • TR 09 02 | - Solve cos(x) = a
  • TR 09 03 | - Solve tan(x) = a
  • TR 09 04 | - Solve sin(x)^2 = 1/4
  • TR 09 05 | - Solve sin(x) + sin(3*x) = 0
  • TR 09 06 | - Solve sin(x) + sin(5*x) = 0
  • TR 09 07 | - Solve x^2 - 6*x + 16*sin(A) = 0
  • TR 09 08 | - Solve 1/cos(x) + 1/sin(x) = 2*Sqr(2)
  • TR 09 09 | - Solve sin(5*x) + sin(11*x) = 0
  • TR 09 10 | - Solve sin(2*x)*sin(5*x) + sin(4*x)*sin(11*x) + sin(9*x)*sin(24*x) = 0
  • TR 09 11 | - Solve sin(x)^3 + cos(x) = sin(x) + cos(x)^3
  • TR 09 12 | - Solve 1 - cos(x) = Sqr(3)*sin(x)
    • Answers


    TR 09 01. Solve sin(x) = a

    General solutions
    • sin(x) = 1/2
      • Since sin(x) is positive, x is 1st or 2nd quadrant.
      • The principal solution in first quadrant is pi/6.
      • Hence x = 2*n*pi + pi/6 and x = 2*(n+1)*pi - pi/6.
    • sin(x) = -1/2
      • Since sin(x) is negative, x is 3rd or 4th quadrant.
      • The principal solution in first quadrant is pi/6.
      • Hence x = 2*n*pi - pi/6 and x = 2*(n+1)*pi - pi/6.
    General solution of sin(x) = 0
    • Since x=0, x=180, x = 360, ....
    • Hence general solution if x = n*pi.

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    TR 09 02. Solve cos(x) = a

    General solutions
    • cos(x) = 1/2
      • Since cos(x) is positive, x is 1st or 4th quadrant.
      • The principal solution in first quadrant is pi/3.
      • Hence x = 2*n*pi + pi/3 and x = 2*n*pi - pi/3.
    • cos(x) = -1/2
      • Since sin(x) is negative, x is 2nd or 3rd quadrant.
      • The principal solution in first quadrant is pi/6.
      • Hence x = 2*(n+1)*pi - pi/6 and x = 2*(n+1)*pi + pi/6.
    General solution of cos(x) = 0
    • Since x=90, x=270, x = 450, ....
    • Hence general solution if x = n*pi + pi/2.

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    TR 09 03. Solve tan(x) = a

    General solutions
    • tan(x) = 1
      • Since tan(x) is positive, x is 1st or 3rd quadrant.
      • The principal solution in first quadrant is pi/4.
      • Hence x = n*pi + pi/4
    • tan(x) = -1
      • Since tan(x) is negative, x is 2nd or 4th quadrant.
      • The principal solution in first quadrant is pi/6.
      • Hence x = n*pi - pi/4.
    General solution of tan(x) = 0
    • Since x=0, x=180, x = 360, ....
    • Hence general solution if x = n*pi.
    Example : Solve tan(A) = -1 for A between -360 and 360
      1. Angle A between -360 and 360
        * Tan(180-45) = -tan(45) = -1 and A = 135
        * Tan(360-45) = -tan(45) = -1 and A = 315
      2. General solution
        * Principal angle A = 45 for tan(A)=1
        * Tan(A) = (-) and A is in 2nd quadrant or 4th quadrant
        * Hence A = 180*k - 045
        * Where k=0,1,2,3,4,... k=even in 4th quadrant and k=odd in 2nd quadrant
    Example : Solve tan(A) = 1
      1. Angle A between -360 and 360
        * Tan(45) = 1 and A = 45
        * Tan(225) = tan(180+45) = tan45 = 1 and A = 225
      2. General solution
        * Principal angle A = 45 for tan(A) = 1
        * Tan(A) = (+) and A = 45 in 1st quadrant and 225 in 3rd quadrant
        * Hence A = 180*k + 045 in 1st and 3rd quadrant
        * Where k=0,1,2,3,4,... k = even in 1st quadrant and k = odd in 3rd quadrant

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    TR 09 04. Solve sin(x)^2 = 1/4

    Hint
    • This is to solve sin(x) = 1/2 and sin(x) = -1/2
    Solution
    • For sin(x) = +1/2 and x = 2*n*pi + pi/6 and x = (2*n + 1)*pi - pi/6
    • For sin(x) = -1/2 and x = 2*n*pi - pi/6 and x = (2*n + 1)*pi + pi/6
    • Hence the general solutions are x = n*pi + pi/6 and x = n*pi - pi/6

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    TR 09 05. Solve sin(x) + sin(3*x) = 0
    Method 1 : Use sin(3*x) formula
    • sin(x) + 3*sin(x) - 4*sin(x)^3 = 0.
    • 4*sin(x) - 4*sin(x)^3 = 0.
    • 4*sin(x)*(1 - sin(x)^2) = 0.
    • sin(x) = 0, hence x = n*pi.
    • 1 - sin(x)^2 = 0
      • sin(x) = +1. General solution is 2*n*pi + pi/2
      • sin(x) = -1. General solution is (2*n+1)*pi + pi/2
      • Hence x = n*pi + pi/2
    Method 2 : use sum of functions
    • 2*sin(x+3x)/2)*cos((3*x-x)/2) = 0.
    • Hence sin(2*x)*cos(x) = 0.
    • For sin(2*x) = 0.
      • General solution 2*x = n*pi.
      • Hence x = n*pi/2.
    • For cos(x) = 0.
      • General solution x = n*pi + pi/2.

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    TR 09 06. Solve sin(x) + sin(5*x) = 0

    Method 1 : use sin(5*x)
    • sin(x) + 5*sin(x) - 20*sin(x)^3 + 10*sin(x)^4 = 0
    • sin(x)*(6 - 20*sin(x)^2 + 10*sin(x)^2) = 0
    • For sin(x) = 0.
      • x = n*pi.
    • For 2*(3 - 10*sin(x)^2 + 5*sin(x)^2) = 0
    • sin(x)^2 = (-(-10) + Sqr((-10)^2 - 4*3*5))/(2*3) = (10 + Sqr(40))/6 = 2.720759.
    • Since sin(x) is between -1 and 1, hence this is not solution.
    • sin(x)^2 = (-(-10) - Sqr((-10)^2 - 4*3*5))/(2*3) = (10 - Sqr(40))/6 = 0.612574.
    • sin(x) = 0.78267 or sin(x) = -0.78267.
    • Since x = arcsin(0.78267) = 51.506 degrees.
    • General solution is n*180 + 51.506 or n*180 - 51.506.
    Method 2 : use sum of two functions
    • 2*sin((5*x+x)/2)*cos((5*x-x)/2) = 0.
    • sin(3*x)*cos(2*x) = 0.
    • For sin(3*x) = 0.
      • 3*x = n*pi or x = n*pi/3.
    • For cos(2*x) = 0.
      • 2*x = n*pi + pi/2 or x = n*pi/2 + pi/4.

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    TR 09 07. Solve x^2 - 6*x + 16*sin(A) = 0

    Questions
    • 1. Find A if equation has two interal roots
    • 2. Find A if equation has real roots
    • 3. Find A if equation has complex roots
    Soution : Find sin(A) if it has integral roots
    • Discrimiant : D = (-6)^2 - 4*1*16*sin(A)
    • Since roots r = (-(-6) + Sqr(D))/2 and s = (-(-6) - Sqr(D))/2
    • Hence Sqr of D must be integer and D > 0
    • D = 36 - 64*sin(A)
      • If Sqr(D) is integer, then D = 1, 4, 9, 25, 36, 49, 64, 81, ...
      • If D is integer, then sin(A) must be -1, -1/2, 0, 1/2 and 1
      • If sin(A) = -1, then D = 100
        • r = (6 + Sqr(100))/2 = (6 + 10)/2 = +8
        • s = (6 - Sqr(100))/2 = (6 - 10)/2 = -2
        • Hence sin(A) = -1 and A = 270 degrees or (2*n-1)*pi + pi/2
      • If sin(A) = -1/2, then D = 36 + 32. Sqr(D) is not integer
        • Hence there are no integral solution
      • If Sin(A) = 0, then D = 36
        • r = (6 + Sqr(36))/2 = (6 + 6)/2 = +6
        • s = (6 - Sqr(36))/2 = (6 - 0)/2 = -0
        • Hence sin(A) = 0 and A = 0 degrees or n*pi
      • If sin(A) = 1/2, then D = 4. Sqr(D) is not integer
        • r = (6 + Sqr(4))/2 = (6 + 2)/2 = +4
        • s = (6 - Sqr(4))/2 = (6 - 2)/2 = +2
        • Hence sin(A) = 0 and A = 0 degrees or n*pi
      • If sin(A) = 1, then D = -32. There is no real roots
    • Hence it has integral roots if sin(A) = -1, 0 or 1/2
    Soution : Find sin(A) if it has real roots
    • D = (-6)^2 - 4*1*16*sin(A) must be greater than 0
    • Case 1 :
      • sin(A) is negative and the roots are real
      • Hence A is between (2*n-1)*pi and 2*n*pi
    • Case 2 :
      • 64*sin(A) is less than 32 or sin(A) LT 32/64
      • If sin(A) = 32/64 = 0.5, then A = 30 or 150 degrees
      • Hence A is between 0 and 30 degrees
      • or A is between 150 and 180 degrees
    Soution : Find sin(A) if it has real roots
    • Since sin(A) is less than zero it has real roots
    • Also sin(A) LE 1/2 grive real roots
    • The roots are complex if sin(A) greater than 1/2
    • Hence A is between 30 degrees and 150 degrees

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    TR 09 08. Solve 1/cos(x) + 1/sin(x) = 2*Sqr(2)

    Solution
    • (sin(x) + cos(x))/Sqr(2) = 2*sin(x)*cos(x)
    • sin(x)*cos(pi/4) + cos(x)*sin(pi/4) = sin(2*x)
    • sin(x + pi/4) = sin(2*x)
    • Hence x + pi/4 = 2*n*pi + 2*x
    • -x = 2*n*pi - pi/4
    • Hence x + pi/4 = (2*n+1)*pi - 2*x
    • 3*x = (2*n+1)*pi - pi/4

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    TR 09 09. Solve sin(5*x) + sin(11*x) = 0

    Solution
    • sin((5*x+11*x)/2)*cos((5*x-11*x)/2) = 0
    • Hence sin(8*x) = 0 and 8*x = n*pi
    • Hence cos(3*x) = 0 and 3*x = n*pi + pi/2

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    TR 09 10. Solve sin(2*x)*sin(5*x) + sin(4*x)*sin(11*x) + sin(9*x)*sin(24*x) = 0

    Solution
    • sin(2*x)*sin(05*x) = (cos(07*x) - cos(03*x))/2
    • sin(4*x)*sin(11*x) = (cos(15*x) - cos(07*x))/2
    • sin(9*x)*sin(24*x) = (cos(33*x) - cos(25*x))/2
    • Add LHS and RSH
    • Equation : LHS = (cos(33*x) - cos(3*x))/2 = 0
    • sin(18*x)*sin(15*x) = 0
    • Sin(18*x) = 0 and x = n*pi/18
    • Sin(15*x) = 0 and x = n*pi/15

    Go to Begin

    Go to Begin

    TR 09 11. Solve Solve sin(x)^3 + cos(x) = sin(x) + cos(x)^3

    Solution
    • Rearrange as sin(x)^3 - cos(x)^3 = sin(x) - cos(x)
    • (sin(x) - cos(x))*(sin(x)^2 + sin(x)*cos(x) + cos(x)^2) - (sin(x) - cos(x)) = 0
    • (sin(x) - cos(x))*(sin(x)^2 + sin(x)*cos(x) + cos(x)^2) - 1) = 0
    • (sin(x) - cos(x))*(1 + sin(x)*cos(x) - 1) = 0
    • (sin(x) - cos(x))*sin(x)*cos(x) = 0
    • Solution 1
      • sin(x) - cos(x) = 0
      • Hence tan(x) = 1
      • Hence x = pi/4
      • General solution : x = 2*n*pi + pi/4 or (2*n+1)*pi + pi/2
    • Solution 2
      • sin(x) = 0 and x = 0 or 180 degrees
      • General solution : x = n*pi
    • Solution 3
      • Since cos(x) = 0 and x = 90 or 270 degrees
      • General solution : x = n*pi + pi/2

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    TR 09 12. Solve 1 - cos(x) = Sqr(3)*sin(x)

    Solution
    Go to Begin

    TR 09 00. Outlines

    General solution of sin(x) = a
    • Principal solution : in 1st quadrant and x = arcsin(a)
    • In 1st quadrant : x = 2*n*pi + arcsin(a)
    • In 2nd quadrant : x = (2*n-1)*pi + arcsin(a)
    General solution of sin(x) = -a
    • Principal solution : in 1st quadrant and x = arcsin(a)
    • In 3rd quadrant : x = (2*n-1)*pi + arcsin(a)
    • In 4th quadrant : x = 2*n*pi - arcsin(a)
    General solution of cos(x) = a
    • Principal solution : in 1st quadrant and x = arccos(a)
    • In 1st quadrant : x = 2*n*pi + arccos(a)
    • In 4th quadrant : x = 2*n*pi - arccos(a)
    General solution of cos(x) = -a
    • Principal solution : in 1st quadrant and x = arccos(a)
    • In 2nd quadrant : x = (2*n-1)*pi - arccos(a)
    • In 3rd quadrant : x = (2*n-1)*pi + arccos(a)
    General solution of tan(x) = a
    • Principal solution : in 1st quadrant and x = arctan(a)
    • In 1st quadrant : x = 2*n*pi + arctan(a)
    • In 3rd quadrant : x = (2*n-1)*pi + arctan(a)
    General solution of tan(x) = -a
    • Principal solution : in 1st quadrant and x = arctan(a)
    • In 2nd quadrant : x = (2*n-1)*pi - arctan(a)
    • In 4th quadrant : x = 2*n*pi - arctan(a)
    General solution of sin(x)^2 = a
    • Principal solution : in 1st quadrant and x = arcsin(Sqr(a))
    • Angles in all 4 quadrants
    • x = n*pi + arcsin(a)
    • x = n*pi - arcsin(a)
    General solution of cos(x)^2 = a
    • Principal solution : in 1st quadrant and x = arccos(Sqr(a))
    • Angles in all 4 quadrants
    • x = n*pi + arccos(a)
    • x = n*pi - arccos(a)
    General solution of tan(x)^2 = a
    • Principal solution : in 1st quadrant and x = arctan(Sqr(a))
    • Angles in all 4 quadrants
    • x = n*pi + arctan(a)
    • x = n*pi - arctan(a)
    Solution of quadratic equation a*x^2 + b*x + c = 0
    • x = (-b + Sqr(b^2 - 4*a*c))/(2*a)
    • x = (-b - Sqr(b^2 - 4*a*c))/(2*a)
    Equation theory of quadratic equation a*x^2 + b*x + c = 0
    • Let roots be r and s
    • r + s = -b/a
    • r*s = c/a

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