Trigonometry
Trigonometric Equations
Homogeneous Equations in sin x and cos x
Homogeneous equations of first degree  a×sin x + b×cos x = 0
Homogeneous equations of second degree  a sin2 x + b sin x · cos x + c cos2 x = 0
The Basic Strategy for Solving Trigonometric Equations
Trigonometric equations examples
Homogeneous trigonometric equations in sin x and cos x
An equation is said to be homogeneous if all its terms are of the same degree.
Homogeneous equations of first degree  a sin x + b cos x = 0
Divide given equation by cos x to obtain
a tan x + b = 0   or   tan x = - b/a   the basic equation
whose solution is     x = tan-1 (- b/a) + k · p    or    x = arctan (- b/a) + k · p,  k Î Z.
Example:  Solve the equation,  - sin x + Ö3 · cos x = 0.
Solution:  Dividing given equation by  - cos x obtained is
tan x = Ö3,     x = tan-1Ö3  + k · p = p/3 + k · p,  k Î Z.
Homogeneous equations of second degree  a sin2 x + b sin x · cos x + c cos2 x = 0
After division of the given equation by cos2 x obtained is quadratic equation
a tan2 x + b tan x + c = 0
which is explained in the previous section.
The next example shows that the equation  a sin2 x + b sin x · cos x + c cos2 x = d  is also homogeneous.
Example:  Find the solution set for the equation,  5 sin2 x + sin x · cos x + 2 cos2 x = 4.
Solution:  Given equation is equivalent to the equation
5 sin2 x + sin x · cos x + 2 cos2 x = 4 · (sin2 x + cos2 x),   since  sin2 x + cos2 x = 1
which, when simplified becomes
sin2 x + sin x · cos x - 2 cos2 x = 0  - the homogeneous equation of the second degree.
 Division by cos2 x gives,
thus,    (tan x)1 = - 2,    x = tan-1(- 2) = - 63°2605 + k · 180°,
(tan x)2 1,      x = tan-1 1 = 45° + k · 180°,  k Î 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:  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