How to Solve Trigonometric Equations with Detailed Solutions - Grade 12

Grade 12 trigonometric equations with detailed solutions are presented along with graphical interpretation.

  1. Find all solutions of the equation
    sin(x) = 1 / 2
    in the interval [0 , 2π) and explain the solutions graphically using a unit circle and the graph of sin (x) - 1/2 in a rectangular coordinate system of axes.

    Solution

    1. sin(x) is positive in quadrant I and II and therefore the given equation has two solutions.
    2. In quadrant I, the solution to sin(x) = 1 / 2 is x = π / 6. (green in unit circle)
    3. In quadrant II, and by symmetry of the unit circle, the solution to sin(x) = 1 / 2 is x = π - π / 6 = 5π/6 (brown in unit circle)

      sat question - graphical solution unit circle question 1


    4. The graphical solutions of the equation sin(x) = 1 / 2 are obtained by first writing the equation with right side equal to zero sin(x) - 1 / 2 = 0. Then graph the left side of the equation and the solutions of the equation are the x - intercepts of the graph of f(x) = sin(x) - 1 / 2 as shown below. It is easy to check that the two x intercepts in the graph below are close to the the analytical solutions found above:π / 6 and 5π/6.

      sat question - graphical solution question 1


  2. Find all solutions of the equation
    sin(x) + cos(x) = 0
    in the interval [0 , 2π) and explain the solutions graphically using the unit circle and the graph of sin (x) + cos(x) in a rectangular coordinate system.

    Solution

    1. The equation may be written as: sin(x) = -cos(x).
    2. Divide both sides of the equation by cos(x): sin(x)/ cos(x) = -1
    3. Use the identity tan(x) = sin(x)/cos(x) to rewrite the equation as: tan(x) = - 1
    4. Solve the equation tan(x) = - 1
    5. tan(x) is negative in quadrant II and IV
    6. Solution in quadrant I: x = 3π/4
    7. By symmetry of the unit circle, the solution in quadrant IV is given by: 3π/4 + π = 7π / 4

      sat question - graphical solution unit circle question 2


    8. The graphical solutions may be approximated by the x intercepts of the graph of g(x) = sin(x) + cos(x). We can easily check that the two solutions found above 3π/4 and 7π/4 are close to the x intercepts of the graph shown below.

      sat question - graphical solution question 2


  3. Find all solutions of the equation:
    sin(πx / 2)cos(π/3) - cos(πx / 2)sin(π/3) = 0


    and check graphically the first few positive solutions.

    Solution

    1. We first use the formula sin(A - B) = sin A cos B - cos A sin B
    2. to write : sin(πx / 2)cos(π/3) - cos(πx / 2)sin(π/3) = sin(πx / 2 - π/3)
    3. rewrite the given equation as follows: sin(πx / 2 - π/3) = 0
    4. Solve the above equation to obtain: πx / 2 - π/3 = k π   where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
    5. Simplify and rewrite the solution as: x = 2 k + 2/3   where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

      sat question - graphical solution question 3


    6. The first 3 positive solutions are: 2/3 , 8/3 and 14/3 corresponding to k =0, 1 and 2. These values are close to the x intercepts of the graph of f(x) = sin(πx / 2)cos(π/3) - cos(πx / 2)sin(π/3) shown above.

  4. Find all solutions of the equations: 2 sin2(x) + cos (x) = 1.

    and check graphically the first few positive solutions.

    Solution

    1. We first use the identity: sin2(x) = 1 - cos2(x)
    2. to rewrite the equation as: 2(1 - cos2(x)) + cos (x) = 1
    3. group like terms and write equation with right side equal to zero: -2 cos2(x) + cos (x) +1 = 0
    4. Let u = cos (x) and rewrite the equation as: -2 u2 + u +1 = 0
    5. Solve for u: u = 1 and u = - 1 / 2
    6. First group of solutions: Solve u = 1 for x: u = cos (x) = 1 :         x1 = 2 k π   where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
    7. Solve u = -1 / 2 for x:
    8. cos(x) is negative in quadrant II and III, hence u = cos (x) = - 1/2 has two more groups of solutions given by
    9. Second group of solutions:        x2 = 2π/3 + 2 k π   where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
    10. Third group of solutions:         x3 = 4π/3 + 2 k π   where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

      sat question - graphical solution question 4


    11. The first 3 positive solutions are obtained by setting k = 0 in all three groups of solutions to obtain: 0, 2π/3 and 4π/3. These values are close to the x intercepts of the graph of g(x) = 2 sin2(x) + cos (x) - 1 shown above.

  5. Find all solutions of the equation: 6 cos2(x / 2) - cos (x) = 4.

    and check graphically the first few positive solutions.

    Solution

    1. We first use the identity: cos(x) = 2 cos2(x / 2) - 1
    2. to rewrite the equation as: 6 cos2(x / 2) - (2 cos2(x / 2) - 1 ) = 4
    3. group like terms and write equation with right side equal to zero: 4 cos2(x / 2) = 3
    4. cos(x / 2) = ~+mn~ √3 / 2

    5. Solve the first equation cos(x / 2) = √3 / 2
      the cosine function is positive in quadrant I and IV hence the two groups of solutions:

      x / 2 = π / 6 + 2 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant I)
      and
      x / 2 = 11π / 6 + 2 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant VI)

      Multiply all terms by 2 to obtain the first two groups of solutions:

      x1 = π / 3 + 4 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
      x2 = 11π / 3 + 4 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

    6. Solve the second equation cos(x / 2) = - √3 / 2
      The cosine function is negative in quadrant II and III hence the two groups of solutions:

      x / 2 = 5 π / 6 + 2 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant II)
      and
      x / 2 = 7 π / 6 + 2 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant III)

      Multiply all terms by 2 to obtain two more groups two of solutions:

      x3 = 5 π / 3 + 4 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
      x4 = 7 π / 3 + 4 k π where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

      sat question - graphical solution question 5


    7. The first 4 positive solutions are obtained by setting k = 0 in all four groups of solutions to obtain: π/3 , 5π/3, 7π/3 and 11π/3. These values are close to the x intercepts of the graph of f(x) = 6 cos2(x / 2) - cos (x) - 4 shown above.

  6. Solve the equation: tan2 ( x) + 2 sec (x) + 1 = 0.

    and check graphically the first few positive solutions.

    Solution

    1. We first use the identity: tan2(x) = sec2(x) - 1
    2. to rewrite the equation as: sec2(x) - 1 + 2 sec (x) + 1 = 0
    3. group like terms and write equation with right side equal to zero: sec2(x) + 2 sec (x) = 0
    4. Factor: sec (x) (sec (x) + 2) = 0
    5. sec (x) = 0 has no solution.
    6. Solve : sec (x) + 2 = 0
    7. Use sec (x) = 1 / cos (x) to rewrite equation as : cos(x) = - 1 / 2

    8. The cosine function is negative in quadrant II and III hence two groups of solutions
      x1 = 2π/3 + 2 k π   where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
      x2 = 4π/3 + 2 k π;   where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

      sat question - graphical solution question 6

    9. The first 2 positive solutions are obtained by setting k = 0 in the two groups of solutions to obtain: 2π/3 and 4π/3. These values are close to the x intercepts of the graph of f(x) = tan2 ( x) + 2 sec (x) + 1 shown above.

  7. Solve the equation: cos(3x) + sin(2x) = cos(x).

    Solution

    1. We first use the identities:      
      cos(3x) = cos3(x) - 3 cos(x)sin2(x)      
      and
            sin(2x) = 2sin(x) cos(x)
      to rewrite the given equation as follows:      cos3(x) - 3 cos(x)sin2(x) + 2sin(x) cos(x) = cos(x)

    2. Rewrite with right hand side equal to zero and factor cos(x)      
            cos(x) ( cos2(x) - 3 sin2(x) + 2 sin(x) - 1) = 0

    3. Use the identity cos2(x) = 1 - sin2(x) to rewrite equation as follows
    4. cos(x) (1 - sin2(x) - 3 sin2(x) + 2 sin(x) - 1) = 0

    5. Simplify
    6. cos(x) (- 4 sin2(x) + 2 sin(x) ) = 0

    7. Factor
    8. 2 sin(x) cos(x) (1 - 2 sin(x)) = 0

    9. Set each factor equal to zero and solve.
    10. sin(x) = 0 gives solutions:           x1 = kπ
    11. cos(x) = 0 gives solutions:            x2 = π/2 + kπ
    12. 1 - 2 sin(x) = 0 or sin(x) = 1/2 gives 2 other gropus of solutions.
                x3 = π/6 + 2kπ
      and
                 x4 = 5π/6 + 2kπ

      sat question - graphical solution question 7


    13. The first 6 positive solutions are obtained by setting k = 0 and k = 1 in the first two groups of solutions x1 and x2 to obtain: 0 , π, π/2, 3π/2, and k = 0 in the second and third groups of solutions x3 and x4 to obtain π/6 and 5π/6. These 6 solutions are close to the x intercepts of the graph of f(x) = cos(3x) + sin(2x) - cos(x) shown above.

  8. Solve the equation: cos(3x) = cos(2x + π/4).

    Solution

    1. We use the fact that if cos(A) = cos(B), then A = B + 2kπ or A = - B + 2kπ to write:

      3 x = 2 x + π/4 + 2kπ  
      or
      3 x = -(2 x + π/4) + 2kπ

    2. Solve 3 x = 2 x + π/4 + 2kπ   Solutions:   x1 = π/4 + 2kπ
    3. Solve 3 x = -(2 x + π/4) + 2kπ ,
      gives 5x = - π/4 + 2kπ
    4. divide all terms by 5 : x2 = - π/20 + 2kπ / 5
      where   k = 0 , ~+mn~ 1 , ~+mn~ 2, ... in both groups of solutions

      sat question - graphical solution question 8


    5. The 6 x intercepts shown in the graph of f(x) = cos(3x) - cos(2x + π/4) below correspond the solutions k = 0 in the first group x1 and k = 0, 1, 2 ,3 and 4 in the second group of solutions x2 = - π/20 + 2kπ / 5.


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