Trigonometry
     Trigonometric Equations
      Equations that Can be Written as  f g = 0
      Trigonometric Equations of Quadratic Form
      The Basic Strategy for Solving Trigonometric Equations
         Trigonometric equations examples
Equations that can be written as  f g = 0
If the given equation, by using appropriate transformations, can be rearranged to the form   f  g = 0 then its solution is represented as the union of the individual solutions of the equations  f = 0 and g = 0.
Example:  Solve the equation,  sin (x + 30) + sin (30 - x) = 2cos2 x.
Solution:  Using the sum to product formula (or addition formula)
given equation gets the form
                                                                  that is,        2sin 30  cos x = 2cos2 x
      or        cos x - 2cos2 x = 0,     cos x  (1 - 2cos x) = 0    the equation of the form  f  g = 0
                                 therefore,       cos x = 0,        x = 90 + k 180,  k Z,
             and     1 - 2cos x = 0,      cos x = 1/2,      x+ 60 + k 360,  k Z.
Thus, the solution set of the given equation we can write as
{90 k 180k Z} U {+ 60 + k 360 ,  k Z} or {90 + k 180+ 60 + k 360 k Z}.
Trigonometric equations of quadratic form
The trigonometric equation of the quadratic form  [F (x)]2 + p F (x) + q = 0, where F (x) denotes given trigonometric function, by substituting  F (x) = u becomes a quadratic equation
returned into substitutions F (x) = u1 and  F (x) = u2  lead to the known basic trigonometric equation.
Example:  Find the solution set for the equation,  3sin x = 2cos2 x.
Solution:  Using known identity we write  3sin x - 2 (1 - sin2 x) = 0,  and by plugging  sin x = u
Therefore,  sin x = 1/2,   x = 30 + k 360,  k Z,  and   x = 150 + k 360,  k Z,
while the equation sin x = - 2  has no solutions since  - 2 is not in the range of the sine function.
Thus, the solution set of the given equation is  {30 + k 360150 + k 360k Z}.
The basic strategy for solving trigonometric equations
When solving trigonometric equations we usually use some of the following procedures,
 - apply known identities to modify given equation to an equivalent expressed in terms of one function,
 - rearrange the given equation using different trigonometric formulae to an equivalent, until you recognize one 
   of the known types.
Trigonometric equations examples
Example:  Solve the equation,  3 sin (x + 70) + 5 sin (x + 160) = 0.
Solution:  Given equation can be written as
4 sin (x + 70) - sin (x + 70) + 4 sin (x + 160) + sin (x + 160) = 0
                                         or    4 [sin (x + 70) + sin (x + 160)] = sin (x + 70) - sin (x + 160)
then, by using sum to product formula
   
                                                         cot (- 45) tan (x + 115) = 1/4   or   tan (x + 115) = - 1/4,
therefore, the solution   x + 115 = tan-1 ,    x = -115 + tan-1 (-1/4) = - 129210 + k 180.
Example:  Find the solution of the equation,  2 sin (x + 60) cos x = 1.
Solution:  Applying products as sums formula
2 (1/2) [sin (x + 60 + x) + sin (x + 60 - x)] = 1   or   sin (2x + 60) + sin  60 = 1
then   sin (2x + 60) = 1 - 3/2,         2x + 60 = sin-1(1 - 3/2) + k 360
                                            and        2x + 60 = 180 - sin-1(1 - 3/2) + k 360
so, the solution is    2x = - 60 + sin-1(1 - 3/2) + k 360,       x = - 2691 + k 180,
                             2x 120 - sin-1(1 - 3/2) + k 360,       x 5691 + k 180,  k Z.
Example:  Find the solution of the equation,
Solution:  Using identities
given equation becomes  2 (1 + cos x) - 3 cot x/2 = 0,
therefore,      1 + cos x = 0,       cos x = - 1,      x = p + k 2p,
       and    2sin x - 3 = 0,      sin x = 3/2,      x = p/3 + k 2p   and   x 2p/3 + k 2p,  k Z.
Trigonometry contents B
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