Trigonometry
     Trigonometric Equations
      Equations of the Type   a cos x + b sin x = c
         Introducing an auxiliary angle method
         Introducing an auxiliary angle method example
         Introducing new unknown  t = tan x/2
         Introducing new unknown  t = tan x/2 example
      The Basic Strategy for Solving Trigonometric Equations
         Trigonometric equations examples
Trigonometric equations of the type   a cos x + b sin x = c
To solve the trigonometric equations which are linear in sin x and cos x, and where, a, b, and c are real numbers we can use the two methods,
a)  introducing an auxiliary angle, and  b)  introducing new unknown.
a)  Introducing an auxiliary angle method to solve trigonometric equation
Consider the constants a and b as rectangular coordinates of a point expressed by polar coordinates (r, j),
then,         a = r cos j   and    b = r sin j,
   
By substituting for a and b in the given equation 
a · cos x + b · sin x = c
obtained is,   r cos x · cos j + r sin x · sin j = or
using addition formula yields, since
obtained is the basic trigonometric equation whose solution is known.
Note that the given equation,  a cos x + b sin x = c  will have a solution if
 
it follows that the constants, a, b and c should satisfy relation  c2 < a2 + b2.
Introducing an auxiliary angle method example
Example:  Solve the equation,  sin x + Ö3 · cos x = 1.
Solution:  Comparing corresponding parameters of the given equation with a cos x + b sin x = c it follows,
a = Ö3, b = 1 and  c = 1.
By substituting given quantities to the basic equation
or  x - 30° = + 60° + k · 360°  thus, solutions are, x = 90° + k · 360° and  x = - 30° + k · 360°, kÎ Z.
The same solution can be obtained using following procedure, from  sin x + Ö3 · cos x = 1
             and   a cos x + b sin x = c | ¸ b
that means that we can introduce an auxiliary angle j that is
sin x · sin 30° + cos x · cos 30° = sin 30°,     cos (x - 30°) = 1/2
and this is the same basic equation obtained above.
b)  Introducing  new unknown  t = tan x/2
If in the equation a cos x + b sin x = c we substitute the sine and cosine functions by tan x/2 = t  that is,
the equation becomes  and after rearranging (a + c) · t2 - 2b · t + (c - a) = 0.
Obtained quadratic equation will have real solutions t1,2 if its discriminant is greater then or equal to zero,  
that is if   (-2b)2 - 4 (a + c)(c - a) > 0   or   c2 < a2 + b2,   which is earlier mentioned condition.
If this condition is satisfied, the solutions, t1 and t2 can be substituted into tan x/2 = t1 and  tan x/2 = t2.
Thus, obtained are the basic trigonometric equations.
Introducing new unknown  t = tan x/2 example
Example:  Solve the equation,  5 sin x - 4 cos x = 3.
Solution:  Given equation is of the form a cos x + b sin x = c therefore parameters are, a = - 4, b = 5 and  
c = 3, after introducing new unknown tan x/2 = t and substituting the values of the parameters into equation
(a + c) · t2 - 2b · t + (c - a) = 0    gives   (- 4 + 3) · t2 - 2 · 5 · t + [3 - (- 4)] = 0
 or      t2 + 10t - 7 = 0,    t1,2  = - 5 + Ö25 + 7 = - 5 +2.
Obtained values for variable t we plug into substitutions,
         tan x/2 = t1,   x/2 = tan-1 (t1 or  x = 2arctan(- 5 - 2)
                                                             x = 2 · (- 84°3821 + k · 180°) = - 169°1642 + k · 360°,
         tan x/2 = t2,   x/2 = tan-1 (t2 or  x = 2arctan(- 5 + 2)
                                                             x = 2 · (33°1756 + k · 180°) = 66°3553 + k · 360°.
The same result we obtain using the method of introducing the auxiliary angle j. Plug the given parameters
a = - 4, b = 5 and c = 3 into  tan j = b/a,  tan j = 5/(- 4)  =>  j = -51°2024,   cos j = 0.624695,  
then from the equation  cos (x + 51°2024″) = (3/- 4) · 0.624695
       or  x + 51°2024 = arccos [(3/- 4) · 0.624695],
thus,  x + 51°2024 = + 117°5618 + k · 360°   =>   x = + 117°5618 - 51°2024 + k · 360°.
Trigonometric equations of the form a cos x + b sin x = c we do not solve using the identity
since that way given equation becomes quadratic with four solutions but only two of them satisfy it.
We will solve the equation from the previous example using this method anyway.
Example:  Solve the equation,  5 sin x - 4 cos x = 3 by substituting 
Solution:  Squaring both sides of an equation may introduce extraneous or redundant (not needed) solutions.
By plugging the results into given equation show that only solutions b) and c) satisfy the equation what match with previous results obtained using another two methods.
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:  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 - Ö3/2,         2x + 60° = sin-1(1 - Ö3/2) + k · 360°
                                            and        2x + 60° = 180° - sin-1(1 - Ö3/2) + k · 360°
so, the solution is    2x = - 60° + sin-1(1 - Ö3/2) + k · 360°,       x = - 26°91 + k · 180°,
                             2x 120° - sin-1(1 - Ö3/2) + k · 360°,       x 56°91 + k · 180°,  k Î Z.
Example:  Find the solution of the equation,
Solution:  Using identities
given equation becomes  2 · (1 + cos x) - Ö3 · cot x/2 = 0,
therefore,      1 + cos x = 0,       cos x = - 1,      x = p + k · 2p,
       and    2sin x - Ö3 = 0,      sin x = Ö3/2,      x = p/3 + k · 2p   and   x 2p/3 + k · 2p,  k Î Z.
Example:  Solve the equation,  cos 2x + cos 6x - cos 8x - 1 = 0.
Solution:  To the first two terms apply the sum to product formula and remaining two terms transform using known identity, thus
 and since    1 + cos 2x = 2cos 2 x   then,     1 + cos 8x = 2cos 2 4x   plugging into the given equation
2cos 4x · cos 2x - 2cos 2 4x = 0    or   2cos 4x · (cos 2x - cos 4x) = 0
 it follows that  4cos 4x · sin 3x · sin x = 0   meaning,   cos 4x = 0,   sin 3x = 0  and/or  sin x = 0.  Since
sin 3x = sin (x + 2x) = sin x · cos 2x + cos x · sin 2x = sin x · (cos 2 x - sin 2 x) + cos x · 2sin xcos x
             = sin x · (1 - 2sin 2 x) + 2sin x · (1 - sin 2 x)  = 3sin x - 4sin 3 x,
solutions of the equation  sin x = 0  are included in the solutions of  sin 3x = 0.
        If  sin x = then,   sin 3x = 3sin x - 4sin 3 x = sin x · (3 - 4sin 2 x) = 0.
Therefore, the solution set of the given equation is the union of the solutions of each of the equations, that is
cos 4x = 0,    4x = + p/2 + k · 2p,     x + p/8 + k · p/2 
                   and           sin 3x = 0,     3x = k · p,       x = k · p/3,   k Î Z.
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