M1410503Part2

M1410503Part2 - (Section 5.3: Solving Trig Equations) 5.32...

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(Section 5.3: Solving Trig Equations) 5.32 Third factor u 1 = 0 u = 1 sin x = 1 x = π 2 + 2 n n integer ( ) Solution set: x x = n , x = 6 + 2 n x = 5 6 + 2 n or x = 2 + 2 n n integer ( ) When gathering groups of solutions, you should check to see if there are any more nice symmetries or periodicities you could exploit. No easy ones are apparent here:
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(Section 5.3: Solving Trig Equations) 5.33 PART E: PYTHAGOREAN IDENTITIES Follow-Up Example Solve: 2sin 3 x + sin x + 3cos 2 x = 3 Solution The cos 2 x seems like the odd man out, but we can make it look more like the powers of sin x in the equation. We often prefer conformity. 3 x + sin x + 3cos 2 x = 3 3 x + sin x + 3 1 sin 2 x ( ) = 3 by a Pythagorean Identity ( ) 3 x + sin x 3sin 2 x = 0 3 x + sin x 3sin 2 x = 0 2 u 3 + u 3 u 2 = 0 Let u = sin x . ( ) 2 u 3 3 u 2 + u = 0 We then proceed as in the previous Example.
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(Section 5.3: Solving Trig Equations) 5.34 PART F: EQUATIONS WITH “MULTIPLE ANGLES” Example Solve: 2sin 4 x ( ) = 3 Solution 2sin 4 x ( ) = 3 Isolate the sin expression. sin 4 x ( ) = θ = 3 2 Substitution: Let = 4 x . sin = 3 2 We will now solve this equation for . Observe that sin π 3 = 3 2 , so 3 will be the reference angle for our solutions for . Since 3 2 is a negative sin value, we want brothers of 3 in Quadrants III and IV. For multiple angle problems, we may prefer positive brothers, as we will see in our Follow-Up Example.
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(Section 5.3: Solving Trig Equations) 5.35 Our solutions for θ are: = 4 π 3 + 2 n , or = 5 3 + 2 n n integer ( ) Note: Our solution set will contain the conjunction “or” as an inclusive (not exclusive or limiting) device to gather solutions together. Although the conjunction “and” may have seemed more appropriate in the above phrasing, we will stick with “or” throughout.
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M1410503Part2 - (Section 5.3: Solving Trig Equations) 5.32...

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