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(2) Trigonometry 10

# (2) Trigonometry 10 - Lesson 1 Solving Equations Pre...

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Solving Equations Lesson 1

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Pre – Calculus Math 40S: Explained! www.math40s.com 136 Trigonometry 2 - Lesson 1 Part One - Graphically Solving Equations Solving trigonometric equations graphically: When a question asks you to solve a system of trigonometric equations, they are looking for the values of θ that make both equations true. There are two ways you can solve for θ : graphically in your TI-83, and algebraically. Part I will show the graphing method, and Parts II & III will focus on algebraic methods. Example 1: Solve 1 cos = 2 θ and state the general solutions: In your TI-83, graph each equation in degree mode. Example 2: Solve 2 cos2 = 2 θ and state the general solutions: Graph both equations in your TI-83, then solve for the first two intersection points. The first two intersection points are at 22.5º and 157.5º. As you can see in the graph, the solutions repeat themselves every period. Since the b -value is 2, the period is 180º, or π . The first general solution is: 22.5º ± n (180º) or 8 n π π ± And the second is: 157.5º ± n (180º) or 7 8 n π π ± Now use 2 nd Æ Trace Æ Intersect to find the points of intersection. They occur at 60º & 300º If you extend the window, you will see that the intersection points are in the same relative places, one period later. The first general solution is: 60º ± n(360º) or (2 ) 3 n π π ± and the second is: 300º ± n(360º) or 5 (2 ) 3 n π π ±
Pre – Calculus Math 40S: Explained! www.math40s.com 137 Trigonometry 2 - Lesson 1 Part One - Graphically Solving Equations Example 3: Graphically find the general solutions for 2sin - 3 θ = 0 Note that even if you manipulate the equation, you can still solve by graphing: If you re-arrange the equation to 2sin = 3 θ by taking 3 to the other side, we get: If we manipulate the equation again by dividing both sides by 2, we get: 3 sin = 2 θ . Solving this: Solving, we still have the same answers of 60º & 120º Graph the two equations in your TI-83 and solve by finding the points of intersection. 60º ± n(360º) or (2 ) 3 n π π ± 120º ± n(360º) or 2 (2 ) 3 n π π ± (In this case, another method would be to find the x-intercepts using 2 nd Æ Trace Æ Zero) Once again, we still get the same answers of 60º & 120º Manipulating an equation does NOT change the solution!

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