Solve Trigonometric Equations

Example 1: Solve the trigonometric equation.(find all solutions)

Solution to example 1

• solve for cos(x)
  
  cos x = 1/2
  
  
• solve for x by finding all values in the interval [0 , 2pi) that statisfy the above trigonometric equation. In this case, with cosine positive and equal to 1 / 2, there are two values: one in the first quadrant of the unit circle.
  
  x1 = pi / 3
  
  
• and a second one in the fourth quadrant (see the two solutions in unit circle in figure below).
  
  x2 = 2*pi - pi / 3 = 5*pi / 3
  
  
  
• find all solutions using the fact that of cos x has a period of 2pi
  
  x1 = pi / 3 + 2*k*pi
  
  x2 = 5*pi / 3 + 2*k*pi
  
  where k is any integer
  
  
  

Example 2: Find the solutions in the interval [0 , 2pi) for the trigonometric equation

Solution to example 2

• change cos ²x to 1 - sin ²x
  
  -5(1 - sin ²x) + 9 sin x = -3
  
  
• mutliply factors and group to obtain
  
  5 sin²x + 9sin x -2 = 0
  
  
• let u = sinx and substitute to obtain an quadratic equation.
  
  
  5 u² + 9 u - 2 = 0
  
  
• use any method to solve for u. By the quadratic formula, we obtain two solutions u1 and u2
  
  u1 = [ -9 - sqrt(121) ] / 10 = 1 / 5 = -2
  
  
• and
  
  u2 = [ -9 + sqrt(121) ] / 10 = 0.2
  
  
• we now solve the equation for x
  
  u1 = sin x = -2
  
  
• the above equation has no solutions for x since -2 is not in the range of values of sin(x);
  -1 <= sin x <= 1.
  
  u2 = sin x = 0.2
  
  
• sin x is positive in the first and second quadrants of the unit circle. There are two solutions to the above equation in [0 , 2pi)(see unit circle below)
  
  solution in the first quadrant
  x1 = arcsin 0.2
  
  
• arcsin 0.2 is the inverse sine function and it can be approximated using a calculator)
  
  solution in the second quadrant
  x2 = pi - arcsin 0.2
  conclusion: There are two solutions to the given equation.
  

Example 3: Find all solutions for the trigonometric equation

Solution to example 3

• subtract cot x from both sides of the equation and simplify
  
  cot x cos ²x - cot x = 0
  
  
• Factor cot x
  
  cot x (cos ²x - 1) = 0
  
  
• Setting each factor in the above trigonometric equation to zero, we obtain two equations.
  
  
  cot x = 0 and cos ²x - 1 = 0
  
  
• The solutions to equation cot x = 0 are given by
  
  x = pi / 2 + k*pi , k is am integer.
  
  
• Equation cos ²x - 1 = 0 gives
  
  cos x = 1 and cos x = -1
  
  
• The solutions to the above equations are given by
  
  x = 2k*pi and x = (2k + 1)*pi where k is an integer
  
  
• HOWEVER the above cannot be solution to the given equation since cot x is undefined for x = 2k*pi and x = (2k + 1)*pi.
  
  conclusion: The solutions to the given equation are.
  
  x = pi / 2 + k*pi where k is an integer.