11.1.
Equations of types cos=
A and sin=
A are called simplest.
11.2.
The trig functions are periodic ones. Therefore,
for infinitely many arguments, they have the same
value.
11.3.
sin
= 2.
11.4
11.5.
1) Factor the equation: sin x(sin x – 1)
= 0. It can be broken into two simplest equations
sin x = 0 and sin x = 1. The solution can be described
by the two sets:
x = n
and x = /2
+ 2n.
2) Factor the equation: (2cos x – 1)(cos
x + 1) = 0. It can be broken into two equations:
cos x = ½ and cos x = –1. The solution
is described by two sets:
x = ± /3
+ 2n
and x =
+ 2n.
3) Factor the equation: (sin x + 2)(sin x –
3) = 0. From here we get two equations:
sin x = –2 and sin x = 3. None of them has
a solution. Therefore,
the original equation does not have solutions.
4) Substitute sin^{2} x = 1 – cos^{2}
x. The equation becomes cos^{2} x –
2cos x – 3 = 0 or
(cos x + 1)(cos x – 3) = 0. Two simplest
equations are cos x = –1 and cos x = 3.
The second equation does not have solutions. The
first equation gives the solutions
x =
+ 2n.
5) The equation can be broken into two equations
sin x = 1 and sin x = –1. The solutions
are x = /2
+ n.
6) The equation can be broken into two equations
tan x = 1 and tan x = –1. The solutions
are x = ± /4
+n.
These solutions can also be written in the form:
x = /4
+ n/2.
7) Factor: sin x(2sin^{2} x – 1)
= 0. From here, sin x = 0 and sin^{2}
x = ½. The first equation gives x
= n.
The second equation gives x = ±
(–1)^{n} /6
+ n.
8) Factor: (2sin^{2} x – 1)( sin^{2}
x – 1) = 0. From here, sin^{2} x
= ½ and sin^{2} x = 1. The first
equation gives x = /4
+ n/2.
The second equation gives x = /2
+ n.
9) Substitute cot x = cos x / sin x. The equation
can be reduced to 2sin^{2} x – 3cos
x = 0. Substitute sin^{2} x = 1 –
cos^{2} x. The equation becomes 2cos^{2}
x + 3cos x – 2 = 0 or (2cos x – 1)(cos
x +2) = 0. From here, cos x = ½ and cos
x = – 2. The second equation does not have
solutions. The first equation gives x
= ± /3
+ 2n.
10) Use the formula sin 2x = 2sin x·cos
x. The equation can be written as
cos x(2sin x – 1) = 0. from here, cos x
= 0 and sin x = ½. The first equation gives
x = /2
+ n.
The second equation gives x = (–1)^{n}/6
+ n.
11.6.
The equation 2sin(3t + 1) = 0 has the solutions
t=(1/3)(n
 1). The first three positive solutions are t
= (–
1)/3, t = (2–
1)/3 and t = (3–
1)/3.
