Trigonometry
Trigonometric Equations
Trigonometric Equations of Quadratic Form
a · cos x + b · sin x = c
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 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°,  kÎ Z,  and   x = 150° + k · 360°,  kÎ 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 · 360°,  150° + k · 360° ,  kÎ Z}.
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.
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 ± 4Ö2.
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°20′24″) = (3/- 4) · 0.624695
 or  x + 51°20′24″ = 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:  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(-1/4) ,    x = -115° + tan-1 (-1/4) = - 129°210 + 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 - 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.
Trigonometry contents B