Trigonometry
Trigonometric Equations
f · g = 0
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°,  kÎ Z,
and     1 - 2cos x = 0,      cos x = 1/2,      x+ 60° + k · 360°,  kÎ Z.
Thus, the solution set of the given equation we can write as
{90° k · 180° ,  kÎ Z} U {+ 60° + k · 360° ,  kÎ Z} or {90° + k · 180°,  + 60° + k · 360° kÎ Z}.
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}.
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) = - 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