Punto de Equilibrio

Trigonometric equations occur whenever trigonometric functions are being anal-ysed, and careful study of them is essential. This section presents a systematic approach to their solution, and begins with the account given in Chapter Four of the Year 11 volume when the compound-angle formulae were not yet available.

Simple Trigonometric Equations: More complicated trigonometric equations eventu-ally reduce to equations like

cosx = −1, or tanx = −√

3, for − 2π ≤ x ≤ 2π,

where there may or may not be a restriction on the domain. The methods here should be familiar by now.

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SIMPLE TRIGONOMETRIC EQUATIONS: If a trigonometric equation involves angles at the boundaries of quadrants, read the solutions off a sketch of the graph.

Otherwise, draw a quadrants diagram, and read the solutions off it.

W^ORKEDE^XERCISE: Solve: (a) cosx = −1 (b) tanx = −√

3, for−2π ≤ x ≤ 2π

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Simple Trigonometric Equations with a Compound Angle: Most troubles are avoided

by substitution for the compound angle. Any given restrictions on the original angle must then be carried through to restrictions on the compound angle.

8 SIMPLE EQUATIONS WITH A COMPOUND ANGLE: Let u be the compound angle. From the given restrictions onx, find the resulting restrictions on u.

WORKEDEXERCISE: Solve sin(3x +⁵₄^π) = 1, for−π ≤ x ≤ π.

Equations Requiring Algebraic Substitutions: If there are powers or reciprocals of the one trigonometric function present, it is usually best to make a substitution for that trigonometric function.

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ALGEBRAIC SUBSTITUTION FOR A TRIGONOMETRIC FUNCTION: Substituteu for the trig-onometric function, solve the resulting algebraic equation, then solve each of the resulting trigonometric equations.

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Equations with More than One Trigonometric Function, but the Same Angle: This is where trigonometric identities come into play.

10 EQUATIONS WITH MORE THAN ONE TRIGONOMETRIC FUNCTION: Trigonometric identities can usually be used to produce an equation in only one trigonometric function.

W^ORKEDE^XERCISE: Solve the equation 2 tanθ = sec θ, for 0^◦≤ θ ≤ 360^◦: (a) using the ratio identities, (b) by squaring both sides.

SOLUTION:

(a) 2 tanθ = sec θ 2 sinθ

cosθ = 1 cosθ sinθ = ¹₂

θ = 30^◦ or 150^◦

(b) Squaring, 4 tan²θ = sec²θ 4 sec²θ − 4 = sec²θ

sec²θ = ⁴₃ cosθ = ¹₂√

3 or −¹₂√ 3

θ = 30^◦, 150^◦, 210^◦ or 330^◦. Checking each solution,θ = 30^◦ or 150^◦. The Dangers of Squaring an Equation: Squaring an equation is to be avoided if

possi-ble, because squaring may introduce extra solutions, as it did in part (b) above.

If an equation does have to be squared, each solution must be checked in the original equation to see whether it is a solution or not. Here are two very simple equations, both purely algebraic, where the effect of squaring can easily be seen.

(a) Suppose that x = 3.

Squaring, x² = 9

so x = 3 or x = −3.

Herex = −3 is a spurious solution.

(b) Suppose that √

x = −5.

Squaring, x = 25.

But √

25 = 5, not −5.

In fact, there are no solutions.

Equations Involving Different Angles: When different angles are involved in the same trigonometric equation, the usual approach is to use compound-angle identities to change all the trigonometric functions to functions of the one angle.

11 EQUATIONS INVOLVING DIFFERENT ANGLES: Use compound-angle identities to change all the trigonometric functions to functions of the one angle.

Frequently such an equation can be solved by more than one method.

W^ORKEDE^XERCISE: Solve cos 2x = 4 sin²x − 14 cos²x, for 0 ≤ x ≤ 2π:

(a) by changing all the angles tox, (b) by changing all the angles to 2x.

SOLUTION:

(a) cos 2x = 4 sin²x − 14 cos²x cos²x − sin²x = 4 sin²x − 14 cos²x

15 cos²x = 5 sin²x tanx =√

3 or −√ 3 x = ^π₃, ²₃^π, ⁴₃^π or ⁵₃^π

(b)cos 2x = 4 sin²x − 14 cos²x

cos 2x = 4(¹₂ −¹₂cos 2x) − 14(¹₂ +¹₂cos 2x) 10 cos 2x = −5

cos 2x = −¹₂

2x = ²₃^π, ⁴₃^π, ⁸₃^π or ¹⁰₃^π x = ^π₃, ²₃^π, ⁴₃^π or ⁵₃^π

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Homogeneous Equations: Equations homogeneous in sinx and cos x were mentioned earlier as a special case of the application of trigonometric identities.

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HOMOGENEOUS EQUATIONS: An equation is called homogeneous in sinx and cos x if the sum of the indices of sinx and cos x in each term is the same.

To solve an equation homogeneous in sinx and cos x, divide through by a power of cosx to produce an equation in tan x alone.

The expansions of sin 2x and cos 2x are homogeneous of degree 2 in sin x and cos x.

Also, 1 = sin²x + cos²x can be regarded as being homogeneous of degree 2.

W^ORKEDE^XERCISE: Solve sin 2x + cos 2x = sin²x + 1, for 0 ≤ x ≤ 2π.

SOLUTION: Expanding, 2 sinx cos x + (cos²x − sin²x) = sin²x + (sin²x + cos²x) 3 sin²x − 2 sin x cos x = 0

÷ cos²x 3 tan²x − 2 tan x = 0

tanx(3 tan x − 2) = 0

tanx = 0 or tan x = ²₃. Hencex = 0, π or 2π, or x =.. 0·588 or 3·730.

The Equations sinx = sin α, cos x = cos α and tan x = tan α: The methods associ-ated with general solutions of trigonometric equations from the last chapter can often be very useful.

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THE GENERAL SOLUTIONS OFsinx = sin α^,cosx = cos α^ANDtanx = tan α:

• If tan x = tan α, then x = nπ + α, where n is an integer.

• If cos x = cos α, then x = 2nπ + α or 2nπ − α, where n is an integer.

• If sin x = sin α, then x = 2nπ + α or (2n + 1)π − α, where n is an integer.

W^ORKEDE^XERCISE: Solve tan 4x = − tan 2x:

(a) using the tan 2θ formula, (b) using solutions of tanα = tan β.

SOLUTION:

(a) tan 4x = − tan 2x Let t = tan 2x.

Then 2t

1− t² =−t 2t = −t + t³ t³− 3t = 0 t(t²− 3) = 0.

(b) tan 4x = − tan 2x.

Since tanθ is an odd function, tan 4x = tan(−2x)

4x = −2x + nπ, where n is an integer 6x = nπ

x = ¹₆nπ.

Hence tan 2x = 0 or tan 2x =√

3 or tan 2x = −√ 3

2x = kπ or ^π₃ +kπ or −^π₃ +kπ, where k is an integer, x = ¹₆nπ, where n is an integer.

Another Approach to Trigonometric Functions of Multiples of 18^◦: In Chapter Four of the Year 11 volume, we used a construction within a pentagon to generate trigono-metric functions of some multiples of 18^◦. Here is another approach through alternative solutions of trigonometric equations.

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W^ORKEDE^XERCISE: Solve sin 3x = cos 2x, for 0^◦≤ x ≤ 360^◦:

(a) graphically, (b) using solutions of cosα = cos β.

Begin solving using compound-angle formulae, and hence find sin 18^◦and sin 54^◦. [Hint: Use the factorisation 4u³− 2u²− 3u + 1 = (u − 1)(4u²+ 2u − 1).] (a) The graphs of the two functions are sketched 1

opposite. They make it clear that there are five solutions, and from the graph, one solution is 90^◦, and the other four are approximately 20^◦, 160^◦, 230^◦and 310^◦. Alternatively, sin 3x = cos 2x

3 sinx − 4 sin³x = 1 − 2 sin²x, using compound-angle identities.

Let u = sin x.

Then 4u³− 2u²− 3u + 1 = 0

(u − 1)(4u²+ 2u − 1) = 0, by the given factorisation.

The quadratic has discriminant 20, so the three solutions of the cubic are u = 1 or u = ¹₄(−1 +√

5 ) each have two solutions.

From part (b), we conclude that sin 18^◦= sin 162^◦= ¹₄(−1 +√ 5 ).

Also, sin 234^◦= sin 306^◦= ¹₄(−1 −√

5 ), so sin 54^◦= ¹₄(1 +√ 5 ).

Note: From these results, the values of all the trigonometric functions at 18^◦, 36^◦, 54^◦ and 72^◦ can be calculated. See the Extension to the following exercise.

Exercise 2D

4. Use the basic trigonometric identities such as sinx

cosx = tanx to solve, for 0 ≤ x ≤ 2π:

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5. Use Pythagorean identities where necessary (such as sin²x + cos²x = 1) and factoring to solve the following for 0^◦≤ α ≤ 360^◦. Give answers to the nearest minute where necessary.

(a) sin²α = sin α 6. Use compound-angle formulae to solve, for 0≤ θ ≤ 2π:

(a) sin(θ +^π₆) = 2 sin(θ −^π₆) (b) cos(θ − ^π₆) = 2 cos(θ + ^π₆)

(c) cos 4θ cos θ + sin 4θ sin θ = ¹₂ (d) cos 3θ = cos 2θ cos θ

[Hint: In part (d), write cos 3θ as cos(2θ + θ).]

7. Use double-angle formulae to solve, for 0≤ x ≤ 2π:

(a) sin 2x = sin x

9. Solve, for 0^◦≤ A ≤ 360^◦, giving solutions correct to the nearest minute where necessary:

(a) 2 sin²A − 5 cos A − 4 = 0

10. Solve, for 0^◦≤ θ ≤ 360^◦, giving solutions correct to the nearest minute where necessary:

(a) 2 sin 2θ + cos θ = 0

(b) Hence solve the homogeneous equation √

3 sin²x + cos²x = (1 +√ 13. Consider the equation cos 3x = cos 2x.

(a) Show that x = ²₅πn, where n is an integer.

(b) Find all solutions in the domain 0≤ x ≤ 2π.

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14. Find the x-coordinates of any stationary points on each of the following curves in the interval 0≤ x ≤ 2π.

(b) Hence use the quadratic formula to solve the equation for 0≤ θ ≤ π.

16. Given the equation 2 cosx − 1 = 2 cos 2x:

(a) Show that cosx = ¹₄(1 +√

5 ) or cosx = ¹₄(1−√ 5 ).

(b) Hence solve the equation for 0≤ x ≤ 360^◦, using the calculator.

17. (a) Show that sin(α + β) sin(α − β) = sin²α − sin²β. 19. Find the values ofk for which:

(a)

(b) Find thex-intercepts for −^π₂ < x < ^3π₂ and the gradient at eachx-intercept.

(c) Show that the curve is concave up at each x-intercept.

(d) Sketch the curve, for−^π₂ < x < ³₂^π.

(b) Hence solve forθ over the given domain, giving solutions to the nearest minute.

[Hint: Beware of the fact that squaring can create invalid solutions.]

25. (a) Show that cos 3x = 4 cos³x − 3 cos x.

(b) By substitutingx = 2 cos θ, show that the equation x³−3x−1 = 0 has roots 2 cos 20^◦,

−2 sin 10^◦ and −2 cos 40^◦.

(c) Use a similar technique to find, correct to three decimal places, the three real roots of the equationx³− 12x = 8√

3 .

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26. (a) If t = tan x, show that tan 4x = 4t(1 − t²) 1− 6t²+t⁴ . (b) If tan 4x tan x = 1, show that 5t⁴− 10t²+ 1 = 0.

(c) Show that sinA sin B = ¹₂(cos(A − B) − cos(A + B)) and that cosA cos B = ¹₂(cos(A − B) + cos(A + B)).

(d) Hence show that ₁₀^π and ^3π₁₀ both satisfy tan 4x tan x = 1.

(e) Hence write down, in trigonometric form, the four real roots of the polynomial equation 5x⁴− 10x²+ 1 = 0.

In document UNIVERSIDAD DE EL SALVADOR FACULTAD DE CIENCIAS ECONÓMICAS (página 115-131)