Solve Trigonometric Equations Questions with Detailed Solutions

Ways of solving trigonometric equations with detailed solutions are presented along with graphical interpretation.

  1. Question 1

    Find all solutions of the equation
    sin(x) = 1 / 2
    in the interval [0 , 2?) 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


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

    sat question - graphical solution unit circle question 1



    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:? / 6 and 5?/6.

    sat question - graphical solution question 1

  2. Question 2

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

    Solution


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

    sat question - graphical solution unit circle question 2



    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 3?/4 and 7?/4 are close to the x intercepts of the graph shown below.

    sat question - graphical solution question 2

  3. Question 3

    Find all solutions of the equation:
    sin(?x / 2)cos(?/3) - cos(?x / 2)sin(?/3) = 0

    and check graphically the first few positive solutions.

    Solution


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

    sat question - graphical solution question 3



    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(?x / 2)cos(?/3) - cos(?x / 2)sin(?/3) shown above.
  4. Question 4

    Find all solutions of the equations: 2 sin2(x) + cos (x) = 1.
    and check graphically the first few positive solutions.

    Solution


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

    sat question - graphical solution question 4



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

  5. Question 5

    Find all solutions of the equation: 6 cos2(x / 2) - cos (x) = 4.
    and check graphically the first few positive solutions.

    Solution


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


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

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

    x1 = ? / 3 + 4 k ? where   k = 0 , 1 , 2, ...
    x2 = 11? / 3 + 4 k ? where   k = 0 , 1 , 2, ...


    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 ? / 6 + 2 k ? where   k = 0 , 1 , 2, ... (quadrant II)
    and
    x / 2 = 7 ? / 6 + 2 k ? where   k = 0 , 1 , 2, ... (quadrant III)

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

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

    sat question - graphical solution question 5



    The first 4 positive solutions are obtained by setting k = 0 in all four groups of solutions to obtain: ?/3 , 5?/3, 7?/3 and 11?/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. Question 6

    Solve the equation: tan2 ( x) + 2 sec (x) + 1 = 0.
    and check graphically the first few positive solutions.

    Solution


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


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

    sat question - graphical solution question 6


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

  7. Question 7

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

    Solution


    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)


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


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


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


    Factor
    2 sin(x) cos(x) (1 - 2 sin(x)) = 0


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

    sat question - graphical solution question 7



    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 , ?, ?/2, 3?/2, and k = 0 in the second and third groups of solutions x3 and x4 to obtain ?/6 and 5?/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. Question 8

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

    Solution


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

    3 x = 2 x + ?/4 + 2k?  
    or
    3 x = -(2 x + ?/4) + 2k?


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

    sat question - graphical solution question 8



    The 6 x intercepts shown in the graph of f(x) = cos(3x) - cos(2x + ?/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 = - ?/20 + 2k? / 5.


References and Links

High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
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