Transcript Solving Trigonometric Equations

Solving Trigonometric Equations
First Degree Trigonometric
Equations:
• These are equations where there is one
kind of trig function in the equation and
that function is raised to the first power.
2 sin( x)  1
Steps for Solving:
• Isolate the Trigonometric function.
• Use exact values to solve and put answers in terms
of radians.
• If the answer is not an exact value, then use inverse
functions on your calculator to get answers
2 sin( x)  1
1
sin( x) 
2
Now figure out where sin = -1/2 on the unit circle.
1
7
11
sin 
at
and
2
6
6
Complete the List of
Solutions:
• If you are not restricted to a specific
interval and are asked to give the general
solutions then remember that adding on
any integer multiple of 2π represents a coterminal angle with the equivalent
trigonometric ratio.
Solutions :
 7

2
k

 6
x
11


 2k
 6
Where k is an integer and gives all
the coterminal angles of the
solution.
Practice
• Solve the equation. Find the general solutions
3 csc  2  0
3 csc  2
2
csc  
3

3
which means that sin 
2
2
   2 k ,
 2 k
3
3
Second Degree Trigonometric
Equations:
• These are equations that have one kind of
Trigonometric function that is squared in
the problem.
• We treat these like quadratic equations
and attempt to factor or we can use the
quadratic formula.
Solve : 4sin 2 ( x)  1  0 over the int erval [0, 2 )
This is a difference of squares and can factor
(2sin x  1)(2sin x  1)  0
Solve each factor and you should end up with 4 solutions
1
1
sin x 
and sin x 
2
2
x
 5 7 11
6
,
6
,
6
,
6
Practice
Find the general solutions for
tan x  2 tan x  1
2
tan x  2 tan x  1  0
2
(tan x  1)(tan x  1)  0
tan x  1
3
7
x
 k ,
 k
4
4
Writing in terms of 1 trig fnc
• If there is more than one trig function
involved in the problem, then use your
identities.
• Replace one of the trig functions with an
identity so there is only one trig function
being used
Solve the following
2 cos x  sin x  1  0
2
Replace cos2 with 1-sin2
2(1  sin 2 x)  sin x  1  0
2  2sin 2 x  sin x  1  0
2sin 2 x  sin x  1  0
2sin 2 x  sin x  1  0
(2sin x  1)(sin x  1)  0
1
sin x 
and sin x  1
2
7
11

x
 2 k , x 
 2 k , x   2 k
6
6
2
Solving for Multiple Angles
• Multiple angle problems will now have a
coefficient on the x, such as sin2x=1
• Solve the same way as previous problems, but
divide answers by the coefficient
• For general solutions divide 2 by the coefficient
for sin and cos. Divide  by the coefficient for
tan and cot.
Find the general solutions for
sin 3x +2= 1
sin3x  1
3
3x 
2
3
2 k
2
x

3
3

2
x   k
2 3
Practice
Solve 2 cos 4 x  3  0
 3
cos 4 x 
2
5
7
4x 
and 4 x 
6
6
5 k
7 k
x

, x

24 2
24 2