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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.
1
)
sin(
2 

x
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
1
)
sin(
2 

x
1
sin( )
2
x


Now figure out where sin = -1/2 on the unit circle.
1 7 11
sin
2 6 6
at and
 


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 co-
terminal angle with the equivalent
trigonometric ratio.














k
k
x
Solutions
2
6
11
2
6
7
:
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
csc 
2
3
2
2 , 2
3 3
k k
 
  
  
3
sin
2
whichmeansthat 
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.
2
: 4sin ( ) 1 0 int [0,2 )
Solve x over the erval 
 
(2sin 1)(2sin 1) 0
x x
  
This is a difference of squares and can factor
Solve each factor and you should end up with 4 solutions
1 1
sin sin
2 2
x and x

 
5 7 11
, , ,
6 6 6 6
x
   

Practice
2
tan 2tan 1
x x
  
2
tan 2tan 1 0
x x
  
(tan 1)(tan 1) 0
x x
  
Find the general solutions for
tan 1
x  
3 7
,
4 4
x k k
 
 
  
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
2cos sin 1 0
x x
  
2
2
2
2
2(1 sin ) sin 1 0
2 2sin sin 1 0
2sin sin 1 0
2sin sin 1 0
x x
x x
x x
x x
   
   
   
  
Replace cos2 with 1-sin2
(2sin 1)(sin 1) 0
1
sin sin 1
2
7 11
2 , 2 , 2
6 6 2
x x
x and x
x k x k x k
  
  
  

 
     
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
sin3 1
x  
3
3
2
x


3
2
2
3 3
k
x


 
sin 3x +2= 1
2
2 3
x k


 
Practice
2cos4 3 0
Solve x  
3
cos4
2
x


5 7
4 4
6 6
x and x
 
 
5 7
,
24 2 24 2
k k
x x
   
   

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Solving_Trigonometric_Equations.advanced function chapter 6ppt

  • 2. 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. 1 ) sin( 2   x
  • 3. 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
  • 4. 1 ) sin( 2   x 1 sin( ) 2 x   Now figure out where sin = -1/2 on the unit circle. 1 7 11 sin 2 6 6 at and    
  • 5. 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 co- terminal angle with the equivalent trigonometric ratio.
  • 6.               k k x Solutions 2 6 11 2 6 7 : Where k is an integer and gives all the coterminal angles of the solution.
  • 7. Practice • Solve the equation. Find the general solutions 3 csc  2  0 3 csc  2 csc  2 3 2 2 , 2 3 3 k k         3 sin 2 whichmeansthat 
  • 8. 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.
  • 9. 2 : 4sin ( ) 1 0 int [0,2 ) Solve x over the erval    (2sin 1)(2sin 1) 0 x x    This is a difference of squares and can factor Solve each factor and you should end up with 4 solutions 1 1 sin sin 2 2 x and x    5 7 11 , , , 6 6 6 6 x     
  • 10. Practice 2 tan 2tan 1 x x    2 tan 2tan 1 0 x x    (tan 1)(tan 1) 0 x x    Find the general solutions for tan 1 x   3 7 , 4 4 x k k       
  • 11. 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
  • 12. Solve the following 2 2cos sin 1 0 x x    2 2 2 2 2(1 sin ) sin 1 0 2 2sin sin 1 0 2sin sin 1 0 2sin sin 1 0 x x x x x x x x                Replace cos2 with 1-sin2 (2sin 1)(sin 1) 0 1 sin sin 1 2 7 11 2 , 2 , 2 6 6 2 x x x and x x k x k x k                  
  • 13. 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.  
  • 14. Find the general solutions for sin3 1 x   3 3 2 x   3 2 2 3 3 k x     sin 3x +2= 1 2 2 3 x k    
  • 15. Practice 2cos4 3 0 Solve x   3 cos4 2 x   5 7 4 4 6 6 x and x     5 7 , 24 2 24 2 k k x x        