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Mathematics Skill Development - Module 7
Mathematics Skill Development - Module 7
Trigonometric Equations
The following questions will evaluate the student’s ability to solve various trigonometric equations.
1. Solve the following equation
sin x = − sin2 x.
Solution:
Rewriting the equation as
sin x = − sin2 x
sin2 x + sin x = 0
sin x(1 + sin x) = 0,
we have that its solutions correspond to sin x = 0 or sin x = −1. In the first case, sin x = 0 when x = kπ
for any integer k ∈ Z. Meanwhile, sin x = −1 when x = 3π
2 + 2kπ for any integer k ∈ Z. Therefore, there is
an infinite amount of number x satisfying this equation characterized by x = kπ and x = 3π
2 + 2kπ where k
could be any integer.
2. Solve the following
tan
x
4
= 1.
Solution:
Since tan(π/4) = 1 and tan(y + π) = tan(y) for any number y, the solutions are given by
x
π
= + kπ, k ∈ Z.
4
4
Therefore, there is an infinite number of solutions to this equation characterized by x = π + 4kπ where k
could be any integer.
3. Solve the following
sin x cos x =
1
.
2
Solution:
We recall the trigonometric identity sin 2x = 2 sin x cos x, to rewrite the left hand side of this equation
as
Therefore, this equation is equivalent to sin(2x) = 1.
Since the solutions to sin y = 1 are given by y = π/2 + 2kπ where k ∈ Z, we obtain 2x = π/2 + 2kπ for any
integer k or
sin(2x)
.
2
x=
π
+ kπ,
4
k ∈ Z.
1
Mathematics Skill Development - Module 7
4. Solve the following
tan2 x cos x + 2 sin x + cos x = 0.
Solution: Dividing this equation by cos x, we obtain a quadratic equation for tan x:
2 sin x
+1=0
cos x
2
tan x + 2 tan x + 1 = 0
tan2 x +
(tan x + 1)2 = 0.
Therefore tan x = −1 or x = −π/4 + kπ for any integer k. Since we divided by cos x in the first step, it
is important to check that the solution we obtain does not correspond to cos x = 0. Since cos π4 = √12 and
√1
cos 3π
4 = − 2 , this is not the case for any of the solutions we obtained.
5. Solve the following
2 cos x − sin x + 2 cos x sin x = 1.
Solution: We simplify the equation by factoring it
2 cos x − sin x + 2 cos x sin x = 1
2 cos x sin x + 2 cos x − sin x − 1 = 0
2 cos x(sin x + 1) − (sin x + 1) − 0
(sin x + 1)(2 cos x − 1) = 0.
Therefore sin x = −1 or cos x = 21 . In the first case, sin x = −1 if x =
x = π3 + 2kπ or x = 5π
3 + 2kπ.
3π
2
+ 2kπ. Meanwhile, cos x =
1
2
if
2