Download 5 - Trig Equations

January 13, 2015
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
5 - Trig Equations
1. Compute the number of values of x for which sin x = 0.7 on the interval 0 ≤ x ≤ 540◦ .
2. Compute all possible integer values of x (measured in degrees) for which sin(x2 − x) = 1.
3. Compute all values of x (measured in radians) over 0 ≤ x < π that solve
3(sin x − cos x) + 4(cos3 x − sin3 x) = 0.
January 2015
5 - Trig Equations
1. Compute the number of values of x for which sin x = 0.7 on the interval 0 ≤ x ≤ 540◦ .
SOLUTION: 4 The equation will have a solution in Quadrant I and a solution in
Quadrant II. Those quadrants are “repeated” in the interval 360◦ ≤ x ≤ 540◦ , so the
answer is 4.
2. Compute all possible integer values of x (measured in degrees) for which sin(x2 − x) = 1.
SOLUTION: 10 + 360n or −9 + 360n must have both We know that sin(90◦ ) = 1, so
we solve x2 − x = 90 → (x − 10)(x + 9) = 0 → x = 10 or x = −9. No other angle whose
sine is 1 produces integer values of x. The answers are 10 or −9 if you’re searching in the
interval 0 ≤ x < 360◦ . However, these will repeat every 360◦ , so the answers are 10 + 360n
or −9 + 360n.
3. Compute all values of x (measured in radians) over 0 ≤ x < π that solve
3(sin x − cos x) + 4(cos3 x − sin3 x) = 0.
SOLUTION:
π 7π 11π
, ,
4 12 12
3
(need all three) Factor the difference of cubes.
3(sin x − cos x) + 4(cos x − sin3 x) = 0 →
−3(cos x − sin x) + 4(cos x − sin x)(cos2 x + cos x sin x + sin2 x) = 0, which factors further to
yield (cos x − sin x)(−3 + 4(1 + cos x sin x)) = 0 → (cos x − sin x)(1 + 4 cos x sin x) = 0 →
(cos x − sin x)(1 + 2 sin 2x) = 0. This implies tan x = 1 → x = π4 or
sin 2x = −1/2 → x = 7π
or x = 11π
. The solution set is { π4 , 7π
, 11π }.
12
12
12 12
January 8, 2014
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
5 - Trig Equations
1. In quadrilateral ABCD, m∠B = 60◦ . If sin A = sin C but A 6= C, compute m∠D in
degrees.
2. Given tan2 x = 2 for 0 ≤ x ≤ 360◦ , compute the sum of the degree measures of the
values of x which satisfy this equation.
3. Compute the number of solutions over 0 ≤ x < 2π: sin 3x + sin
5x + sin7x =0
A−B
A+B
cos
You may find the following formula helpful: sin A + sin B = 2 sin
2
2
January 2014
5 - Trig Equations
1. In quadrilateral ABCD, m∠B = 60◦ . If sin A = sin C but A 6= C, compute m∠D in
degrees.
SOLUTION: 120 Since sin A = sin C but A 6= C, A and C are supplementary.
m∠D = 360 − 180 − 60 = 120.
2. Given tan2 x = 2 for 0 ≤ x ≤ 360◦ , compute the sum of the degree measures of the
values of x which satisfy this equation.
√
SOLUTION: 720 We have that tan x = ± 2. There is a reference angle in the first
quadrant that satisfies this equation, and it has a “mirror image” in Quadrant IV such that
the two angles add to 360◦ . Similarly, there are “mirror image” angles in Quadrants II and
III that add to 360◦ . The total is 720.
3. Compute the number of solutions over 0 ≤ x < 2π: sin 3x + sin
5x + sin7x =0
A−B
A+B
cos
You may find the following formula helpful: sin A + sin B = 2 sin
2
2
A+B
A−B
SOLUTION: 14 We use the identity sin A + sin B = 2 sin
cos
.
2
2
Thus, sin 3x + sin 7x = 2 sin 5x cos 2x, and
sin 3x + sin 5x + sin 7x = 0 → sin 5x(2 cos 2x + 1) = 0. If sin 5x = 0, x = 0 + nπ
and as n
5
assumes values from 0 to 9, we obtain 10 distinct solutions. If 2 cos 2x + 1 = 0, then
cos 2x = −1/2 and x = π6 + mπ or x = 5π
+ mπ, and as m assumes values of 0 or 1 we
6
obtain 4 more solutions for a total of 14 solutions.
January 9, 2013
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
5 - Trig Equations
1. Compute the values of x in degrees over −90◦ < x < 0◦ that satisfy the following
equation: sin(3x) = − 21
2. Compute the values of x in degrees over 0◦ ≤ x < 360◦ that satisfy the following
equation: 3(cot x + csc x) = 2 sin x
3. Compute the values of x in degrees over 0◦ < x < 180◦ that satisfy the following
equation: cos(2x) − (1 − sin2 x) < −0.5
January 2013
5 - Trig Equations
1. Compute the values of x in degrees over −90◦ < x < 0◦ that satisfy the following
equation: sin(3x) = − 21
SOLUTION: −10, −50 x = 330◦ + 360n or 210◦ + 360n ⇒ x = 110◦ + 120n or
70◦ + 120n. Since n = −1, x = −10 or x = −50.
2. Compute the values of x in degrees over 0◦ ≤ x < 360◦ that satisfy the following
equation: 3(cot x + csc x) = 2 sin x
SOLUTION: 120, 240 3(cot x + csc x) = 3 cossinx+1
= 2 sin x for x 6= 0◦ + 180n. This
x
implies
3 cos x + 3 = 2 sin2 x = 2 − 2 cos2 x ⇒ 2 csc2 x + 3 cos x + 1 = (2 cos x + 1)(cos x + 1) = 0.
This solves to give us cos x = −1/2 or cos x = −1, so x = 120◦ or x = 240◦ .
3. Compute the values of x in degrees over 0◦ < x < 180◦ that satisfy the following
equation: cos(2x) − (1 − sin2 x) < −0.5
SOLUTION: 45◦ < x < 135◦ (cos2 x − sin2 x) − 1 + sin2 x < −0.5 ⇒ cos2 x − 1 <√−0.5,
so sin2 x > 0.5. Since sin x is always positive on the given interval, we have sin x > 22 .
Appealing to the graph of y = sin x, we have 45◦ < x < 135◦ .
February 2, 2012
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
3 - Trig Equations
1. If A is an acute angle and sin A = cos 56◦ , compute the number of degrees in angle A.
2. The positive solutions of tan(5x) = 1 are arranged in a list in increasing order. The
fourth value in the list is k ◦ . Compute k.
3. Compute all solutions for x on 0 ≤ x < π: (sin 2x)(cos4 x − sin4 x) = 0
February 2012
3 - Trig Equations
1. If A is an acute angle and sin A = cos 56◦ , compute the number of degrees in angle A.
SOLUTION: 34 We know that sin A = cos(90◦ − A), so we solve A + 56 = 90 to obtain A = 34.
2. The positive solutions of tan(5x) = 1 are arranged in a list in increasing order. The fourth
value in the list is k ◦ . Compute k.
SOLUTION: 117 We have that 5x = 45◦ + 180n, so x = 9◦ + 36n, so the list is
9◦ , 45◦ , 81◦ , 117◦ , · · · . Therefore, k = 117.
3. Compute all solutions for x on 0 ≤ x < π: (sin 2x)(cos4 x − sin4 x) = 0
SOLUTION: {0, π4 , π2 , 3π
4 }(in any order) We factor to obtain
(sin 2x)(cos2 x − sin2 x)(cos2 x + sin2 x) = 0, which simplifies to (sin 2x)(cos 2x) = 0. Using the
sine of the double angle, this may be rewritten as 21 sin 4x = 0. Thus, 4x = 0 + n · π, so
x ∈ {0, π4 , π2 , 3π
4 } (in any order).
FEBRUARY 10, 2010
CALCULATORS MAY NOT BE USED ON ANY TOPICS
#3 TRIG EQUATIONS
ANSWERS:
π
1.
3
2.
105˚, 165˚, 285˚, 345˚
3.
45˚, 150˚, 225˚, 330˚
1. Find arc sin (cos (arc csc 2)), and express your answer in radians.
2. Solve for θ, where 0 < θ ≤ 360 :
(
2 cos θ − 2 sin θ
)
2
=3
[Hint: sin 2θ = 2 sin θ cos θ]
3. Solve for θ, where 0 < θ ≤ 360 :
[Hint: 1 + tan2 θ = sec2 θ]
3 tan 2 θ + tan θ = 3 tan θ + sec2 θ − tan 2 θ
FEBRUARY 3, 2009
CALCULATORS MAY NOT BE USED ON ANY TOPICS
#3 TRIG EQUATIONS
ANSWERS:
1.
4
2.
210˚, 330˚
π π 5π 5π
3.
, , ,
4 3 4 3
1. How many values of x, 0 ≤ x < 360 , satisfy the equation: tan 2 x = tan x ?
2. Solve for θ : 0 < θ ≤ 360
2cos2 θ + 3sin θ = 0
3. Solve for x over 0 ≤ x < 2π:
2 ( cos x − sin x ) = 1 − tan x