Trigonometric equations of quadratic form, Introducing an auxiliary angle method

Trigonometry

Trigonometric equations

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.

a) Introducing an auxiliary angle method

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 = c or

using addition formula yields,

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, 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 t_1,2 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, a = - 4, b = 5 and c = 3, after introducing new unknown tan x/2 = t and substituting 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, t_1,2 = - 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 - 4Ö2)

x = 2 (- 84°38′21″ + k · 180°) = - 169°16′42″ + k · 360°,

tan x/2 = t₂, x/2 = tan^-1 (t₂) or x′ = 2arctan(- 5 + 4Ö2)

x′ = 2 · (33°17′56″ + k · 180°) = 66°35′53″ + k · 360°.

The same result we obtain using the method of introducing an auxiliary angle j. Plug given parameters a = - 4, b = 5 and c = 3 into tan j = b/a, tan j = 5/(- 4), j = -51°20′24″, cos j = 0.624695.

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