Download NAME______________________________________DATE______________________ ALGEBRA2 HONORS MRS. BINASO

NAME______________________________________DATE______________________
ALGEBRA2 HONORS
MRS. BINASO
This assignment is to be completed on loose leaf with all work and formulas shown. There
will be a quiz on Tuesday May 5. If you have any questions as you are working on this,
please feel free to email me.
Solve each of the following equations for all values of  , to the nearest degree, in the
interval 0   360 .
Examples like numbers 1-10 have been completed in class. If assistance is needed refer to
class notes or the homework assignment that was assigned on Thursday April 23rd.
Reminder:
To solve equations with reciprocal functions: change to the suitable function and set it
equal to the reciprocal of the number
3
EX 1: csc  =
EX 2: cot  = .4561
2
2
1
sin  =
tan  =
3
.4561
1. 6(tan  + 2) = 6
2. 6sin  – 3 3 = 0
3. 5sec  + 4 = 3sec 
4. cos2  + 7cos  – 8 = 0
5. tan2  + tan  = 12
6. 2sin2  + 20 = 22sin 
7. sin  cos  + cos  = 0
8. 2sin2  + 4sin  = 0
9. cot2  = 7cot  – 10
10. csc2  + 18 = 3csc 
CASE II FACTORING
Example: 6sin2  – 5sin  – 4 = 0
(6sin2  – 8sin  ) + (3sin  – 4) = 0
2sin  (3sin  – 4) + (3sin  – 4) = 0
(2sin  + 1)(3sin  – 4) = 0
2sin  + 1 = 0
3sin  – 4 = 0
1
4
sin  = 
sin  =
2
3
ref< = 30˚
no solution (sin  cannot be
QIII
QIV
greater than 1)
 = 180˚ +ref<
 = 360˚ – ref<
 = 180˚ + 30˚
 = 360˚ – 30˚
 = 210˚
 = 330˚
 = {210˚,330˚}
The example shown uses factoring by grouping. You may use any Case II method.
11. 2tan2  - 3tan  – 5 = 0
12. 12cos2  – 3 = 5cos 
13. 5sin2  + 26sin  + 5 = 0
14. 8tan2  = 22tan  – 15
15. 3sec2  – 11sec  = 4
16. 10csc2  + 21csc  + 8 = 0
NON-FACTORABLE QUADRATIC EQUATIONS
If a quadratic equation cannot be factored, the quadratic formula must be used.
b  b 2  4ac
x
2a
The difference is that the variable is not x in trigonometric equations so depending on which
function the equation is based the quadratic formula may be:
sin  
b  b2  4ac
2a
OR tan  
OR etc….
b  b2  4ac
b  b2  4ac
OR sec  
2a
2a
(there are 6 trig functions so 6 possibilities)
Example:
3cos  + 4cos  = 6
3cos  + 4cos  – 6 = 0
a=3
b=4
c = -6
cos 
b  b2  4ac
2a
cos 
4  4 2  4(3)( 6)
2(3)
cos 
4  16  72
6
cos 
4  88
6
cos  2.2301
no solution
cos .8968
ref< = 26˚
QI
 = ref<
 = 26˚
QIV
 = 360˚ – ref<
 = 360˚ – 26˚
 = 334˚
 = {26˚, 334˚}
[There could possibly be 4 solutions]
17. 4sin2  + 3sin  – 7 = 0
18. tan2  + 11 = 7tan 
19. 2cos2  = 5cos  + 1
20. 3sec2  + 9sec  = 5