MHF 4U1 Trigonometric Equations

MHF 4U1
Trigonometric Equations
1. Solve for x, Ox2,r.
a) sinx=—-2
b) cosx=—
2
c) tanx=—1
d) secx
e) sinx
2x =
0 cos
=
—2
=
—
—
2. Solveforx,—2rxO.
a) cosx
b) tan
x
2
=
d) sin2x=
=
tanx
c) sin
2 x—sinx
=
2
2 =9
f) (2cscx—l)
x—3=O
2
e) 4cos
3. Solveforx,Ox22T.
a)
cos(2x) + cos x + 1
b) cos(2x) = sin x
c) 3 tan x = tan(2x)
g) sin(2x)+ cos(2x)= 0
e) sin(2x)cosx + sin
2x=1
h) 3cos(2x)+ 2 + cosx = 0
sin(2x)+ sinx = 0
i) sin(2x)= tanx
j)
k) 3sin(2x)+cosx = 0
1) 3sinx+cos(2x)= 2
=
0
d) sinx = 6sin(2x)
2 x—sin
cos
2x
=
2sinxcosx
4. Solve for x in the given interval.
a)
sinx—sinxtanx=O, Oxr
b) cosxtan(3x)=0, —rx0
c) ósin
x—5cosx—2=0, x
2
2
2
e) 2
x—3sin
cos
x
=1, —2nx2n
d) sinx+tanx=O, —x
0
2tanx=secx, —2rx0
5. Solve for x in the given interval.
a)
b) sin(2x)=cosx, Z x
cos(2x)=cos
x
2
, —r x ,r
c) cos
2x
—
2sinxcosx
—
2x
sin
=
0, 0
x
d) 2(sin x + cos
4 x)= 1,
—
,r
x
r
6. Solve for x in the given interval.
a)
cosx=, —2x2
b) sin(2x)=
c) 2
sin
x
+sinx=O, x
d) 2sin
x+sinx—1=0, 0x4
2
e) 2sec
x+3secx—2=0, 0x2
2
0
tanx+sec(2x)=1, x
x
-
—
3
f,’
0
1
l”43
,.
.
hi’
4
j.
)\
n
tJ
Hi
HI
)c,
0
Ij
ct
I
“4
0
I
I,’
ui
VJ
I.
4 H
‘ii
Lj
4%
0
0
0
4
4-
4
I 4
IL
0
A
-‘
‘
cJ
—
‘4ft
1%
0
1
t
.L
-
‘-.
(A
-
I
44
<
--
‘4,
-
(4
x
L
-
I’
I’
‘I
F’)
p
0
‘$1
>p
1,
‘-4
‘I
Y.
‘1
1-
1•
sJ
-ó
—
-A
\s
0
C
0
‘a
—
I
C’
J
0
.4-)
0
o
‘a
L
—I”
0
0
c
0
c
s’
4%
3
‘
(
0
1
—
,L
(A
‘
o
+
c4
41
(
I..
1*
a
a
a
C
0
L
,
.1
4
(t(
I,
4
I’
0
0
4
0
c
41
0
Lf
p
It
ii—’
c)
(
CA
Ii
3
0
N
0
—
I
4..
00
(4
4
L
0
4..
0
0
i
S
1
ii
I
1’
0
0
c
,c
‘
)
3
N
‘I
r)
v-I
-4-
c%J
1
k
‘..
(fi
4
(1
7
4’
0
r\’m
I
‘L
.4L
-p
J
‘4’
‘‘
N
0
0’
1)
‘
L
‘,
.‘j’
•*‘‘
.
‘
‘
.4.
,
a
‘4
‘S
.
L
..
\
.,T
4,
‘,.,
.1
.4.
0
L.
V
c:fr
I”
(C%J
0
0
7
a
0
1’
P
-I
—I- N
I
I
0
1’
0
,1
7
0
I,
tir’
F
0
p
r
7’
)_
-%
I,
‘1
7c
1
0
P
‘I’
“3
Co
(, S
-
I
4%J
t
t
L.
or
I
3(111
-O
0
C)
(rt’)e
or
0
X: t’’()
:r
>:
D.
)L
c)r
Qf
iii
St’’ t
7,.3o
,)
o
‘.
—
I)
—
0
7
0
11
—
I
1,
‘7S
7,
‘1
0
—
0
0
i4
1
4.-,
‘I
I
7-
‘I)
‘p
‘I,
0
S.
1
7;
“1
V
I,
4-
‘JJ
U’
‘I’
tJ
I
+
0f
,€—.
4-
‘0
4(
0
i,
0
1
0
00
/N
o
I
.0
I.
1
N
t1
o
c.
0
-
0
0
1
c
0
c
0
A
S
0
-J
1
1
1
c.
j
.4
‘4
L
3
3
(
0
4
‘F’
ii
0
0
“1
c
1r
a
&
0
1
‘-I
C,’
I
0
0’
I
Cd
‘I’
1
0
4
(I’
L.
L.
0
I
L
L
1
0
(%J
0
—4,,
1
S
1
1’
I
0
,-
I
1.
(4
.f
g
a1.
0
a
&
4
4JS
‘4’
‘
Jb
x
0
0
C’
IL
L
0
iiJ
4
V.
‘\i
L
¶1’
‘4’
c:r
0
I’
t
43
—-——--—
“
‘
—‘
LJ
i’
‘—
‘
.4
4%)
e
C44
%!
4’
41
ci
0
-J
3
4,
0
‘1
1
1’
‘1
9:1
1,
s
1
;
.4
:
)c
\-J
\
*
H-
“
0
-
,
—
w-.
%%J
1,
1.
ç
‘
1
__
,.-,
‘ii
I’
‘I
‘4’
‘4
(
0
L
0
L1%9
C’
I
L
0
4’
0
‘a
i:I%J
‘.-
.,“
4-
Li
vi
‘%_
—
(
0’
a
vi
C’))
I
a
4’
L
0
1
I
Ø1
4-
(i’
(
p
‘U.
(
I’
1
C,,
-4
41
+
3’
:‘D
tcJ
‘I.
U
C’
-
4
_ir
-1
‘-I
0
4-
1
Cd
44
6
L
-4---
‘7,
LI,
q)
‘I
‘I,
4
.
F
c
“3
0
.1
1
13
v)
,Th
-,_ 3:
-
--
—V
—
,
Li)
I-)
0
C’
(
-w
‘1
0
r
x
‘I,
1
7,
x
7’,
‘/1
.
I
4.
I
31
9
LI
I
11
1
7.
C’
0
÷
7L
‘0
(-I,
..
(4
LA
C-’
‘4
*1
‘4
‘4
I
—&
.4
‘1
‘3
(A
t1
44
‘-S
.4-
4
C
a
x
t
I
0
•1
C’
b
c’1
‘3’
‘9
ii
cJ
4-
.7’
4-”
0
p.
I
(1’
4-
r.c3
0
V.’
I
ti
0
14.
(.
I
‘4