Trigonometry
Trigonometric Equations
The Basic Strategy for Solving Trigonometric Equations
Trigonometric equations examples
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.
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.
Trigonometry contents B