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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